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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Light of intensity l is incident perpendicularly on a perfectly reflecting plate of area A kept in a gravity free space. If the photons strike the plate symmetrically and initially the springs were at natural lengths, what is the maximum compression A. `(IA)/(Kc)`B. `(2Ia)/(3Kc)`C. `(3Ia)/(Kc)`D. `(4Ia)/(3Kc)` |
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Answer» Correct Answer - D `(2IA)/(c )=F` and `(K_(eq))` parallel `=3K` `triangleX=(2F)/(3K)=(4IA)/(2Kc)` |
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| 2. |
A. How many photons of a tradiation of wavelength `lamda=5xx10^(-7)` m must fall per second on a blackened plate in order to produce a force of `6.62xx10^(-5)`N? B. At what rate will the temperature of plate rise if its mass is 19.86kg and specific heat is equal to `2500J(kgK^(-1))`?A. `3xx10^(19)`B. `5xx10^(22)`C. `2xx10^(22)`D. `1.67xx10^(18)` |
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Answer» Correct Answer - B The momentum of photon `=(h)/(lamda)` If n is the number of photons falling per second on the plate, then total momentum per second of the incident photons is `P=nxx(h)/(lamda)` Since the plate is blackened, all photons are absorbed by it. `(triangleP)/(trianglet)=n(h)/(lamda)` Since `F=(triangleP)/(trianglet)=n(h)/(lamda)` `n=(Flamda)/(h)` `=(6.62xx10^(-5)xx5xx10^(-7))/(6.62xx10^(-34))=5xx10^(22)` |
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| 3. |
A. How many photons of a tradiation of wavelength `lamda=5xx10^(-7)` m must fall per second on a blackened plate in order to produce a force of `6.62xx10^(-5)`N? B. At what rate will the temperature of plate rise if its mass is 19.86kg and specific heat is equal to `2500J(kgK^(-1))`? |
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Answer» A. If n is the number of photons falling per second on the plate, then the total momentum per second of the incident photons is `P=nxx(h)/(lamda)`. Since the plate is blackened, all photons are absorbed by it. `(triangleP)/(trianglet)=n(h)/(lamda)` Since `F=(triangleP)/(trianglet)=n(h)/(lamda)impliesn=(F.lamda)/(h)` or `n=(6.62xx10^(-5)xx5xx10^(-7))/(6.62xx10^(-34))=5xx10^(22)` B. Energy of each photon `=(hc)/(lamda)` Since n photons fall on the plate per second, the total energy absorbed by the plates in one second is `E=nxx(hc)/(lamda)=1986Js^(-1)` i.e., `(dQ)/(dt)=1986Js^(-1)` `mc(dT)/(dt)=1986` `implies(dT)/(dt)=(1986)/(19.86xx2500)=4xx10^(-2) C``s^(-1)` |
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| 4. |
The frequency and intensity of a light source are both doubled. Consider the following statements A. The saturation photocurrent remains almost the same B. The maximum kinetic energy of the photoelectrons is doubleA. Both A and B are trueB. A is true but B is falseC. A is false but B is trueD. Both A and B are false |
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Answer» Correct Answer - B Here intensity is doubled due to frequency and not by number of photons hence number of ejected electrons remain same. Therefore saturation current remains same. Maximum K.E. will be more than double. |
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| 5. |
Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface.(Mass of the electron `9.11xx10^(-31)kg`)` 1. Calculate the value of work function is eV.A. 1.77B. 3.27C. 5.69D. 2.32 |
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Answer» Correct Answer - B `lamda_0=3800A` `W=hf_0=h(c)/(lamda_0)=(6.633xx10^(-34)xx3xx10^(8))/(3800xx10^(-10))` `=5.23xx10^(-19)J=3.27eV` Incident wavelength `lamda=2600A` `f=` incident frequency `=(3xx10^(8))/(2600xx10^(-10))Hz` Then, `KE_(max)=hf-W-0` `hf=(6.63xx10^(-34)xx3xx10^(8))/(2600xx10^(-10))` `=7.65xx10^(-19)J=4.78eV` `KE_(max)=hf-W-0=4.78eV-3.27eV=1.51eV` `KE_(max)=(1)/(2)mv_(max)^(2)` or `v_(max)=sqrt((2KE_(max))/(m))` |
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| 6. |
In question 23, if the intensity of light is made 4 I, the saturation current will becomeA. `0.4 mu A`B. `0.4xx2 mu A`C. `1.6 mu A`D. `0.4xx16 mu A` |
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Answer» Correct Answer - C Saturation current `prop` intensity `(i_(2))/(0.40) =(4 I)/(I) =4` `(i_(2))/(i_(1)) =4` `implies i_(2) =1.6 mu A` |
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| 7. |
In the previous question, if the intensity of light is made `4I_0`, then the stopping potential will becomeA. `1.36xx1V`B. `1.36xx2V`C. `1.36xx3V`D. `1.36xx4V` |
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Answer» Correct Answer - A Change in intensity from `I_0` to `4I_0` does not affect the stopping potential. |
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| 8. |
(a) Photoelectric threshold of metallic silver is `lambda =3800 Å`. Ultravilolet light of `lambda =260` nm is incident on silver surface. Calculate : (i) the value of work function in joule and eV `K_(max)` of the emitted photoelectrons (iii) `v_(max)` of the photoelectrons (b) Ultraviolet light of wavelength 280 nm is used in an experiment on photoelectric effict with lithium `(phi =2.5 eV)` cathode. Find (i) the maximum kinetic energy of the photoelectrons (ii) the stopping potential (c) Find the maximum magnitude of the linear momentum of a photelectron emitted when light of wavelength 400 nm falls on a metal `(phi =2.5 eV)`. (d) A monochromitc light source of intensity 5 mW emits `8xx10^(5)` photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this set up is `2.0 V` . Calculate the work function of the metal. (e) The maximum kinetic energy of photoelectrons emitted from a certain metallic surface is 30 eV when imonochromatic radiation of wavelength `lambda` falls on it. When the same surface is illuminated with light of wavelength `2 lambda` the maximum kinetic energy of photo electrons is observed to be 10 eV. Calculate the wavelength `lambda` and determine the maximum wavelength of incident radiation for which photoelectrons can be emitted by this surface. |
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Answer» (a) `lambda_(0) =3800 Å =380 nm, lambda =260 nm` (i) `phi =(1242)/(lambda_(0)(nm))eV =(1242)/(380) =3.3 eV` `=3.3xx16xx10^(-19) =5.3xx10^(-19) J` (ii) `E =(1242)/(lambda(nm))eV =(1242)/(260) =4.8 eV` `E=phi + K_(max) implies 4.8 =3.3 + K_(max)` `K_(max) =1.5 eV` (iii) `K_(max) =1.5 eV =1.5xx1.6xx10^(-19) J` `K_(max) =(1)/(2)mv_(max)^(2)` `v_(max) =sqrt((2K_(max))/(m)) =sqrt((2xx1.5xx1.6xx10^(19))/(9.1xx10^(-31)))` `=0.73xx10^(6) m//sec` (b) `E =(1242)/(lambda(nm))eV =(1242)/(280) =4.48 eV` `E = phi + K_(max)` `implies 4.4 =2.5 + K_(max)` `K_(max) =1.9 eV` `K_(max) =eV_(s) implies V_(s) =1.9 V` (c) `E=(1242)/(400) =3.1 eV, phi =2.5 eV` `E =phi + K_(max) implies 3.1 =2.5 + K_(max)` `K_(max) =0.6 eV` `P= sqrt(2mK_(max)) =sqrt(2xx9.1xx10^(-31)xx0.6xx1.6xx10^(-19))` `=4.2xx10^(-25) kg m//sec` (d) `P= n(hc)/(lambda)` `(hc)/(lambda) =(P)/(n) =(5xx10^(-3))/(8xx10^(15))=0.625xx10^(-18) J` `=(0.625xx10^(-18))/(1.6xx10^(-19)) =3.9 eV =E` `K_(max) =eV_(s) =2 eV` `E=phi + K_(max) implies 3.9 = phi + 2` `phi =1.9 eV` (e) `(hc)/(lambda) =phi +30` `(hc)/(2 lambda) =phi +10` (i)/(ii) `2=(phi+30)/(phi+20) implies 2phi + 20=phi +30 implies phi =10 eV` `(hc)/(lambda) =40 eV` `lambda =(1242)/(40) =31 nm` `lambda_(0) =(1242)/(phi(eV))nm =(1242)/(10) =124.2 nm` |
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| 9. |
A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light as shown in the Fig. The mass of the mirror is 20 g. Assume that there is no absorption in in the lens and that `30%` of the light emitted by the source goes through the lens. Find the power (in`xx10^(8)`W) of the source needed to support the weight of the mirror. Take `g=10ms^(-2)` |
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Answer» Let n photons (each of frequency f) per second are emitted from source. Then power of source is `P=nhf` But on `30%` of the photons go towards mirror. The force exerted on mirror is `F=2[(30)/(100)n](h)/(lamda)=(3)/(5)(nhf)/(c)=(3)/(5)(P)/(c)` and this force should be equal weight of mirror, so `(3)/(5)(P)/(c)=20xx10^(-3)g` `impliesP=(5xx3xx10^(8)xx20xx10^(-3)xx10)/(3)=10^(8)W` |
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| 10. |
Some questions (Assertion-Reason Type) are given below. Each question contains Statement I (Assertion) and statement II(reason). Each question has 4 choices (a),(b),(c ) and (d) out of which only one is correct. So select the correct choise. a. Statement I is True, Statement II is True,Statement II is a correct explanation for Statement I b. Statement I is True, Statement II is True, Statement II is NOT a correct ecplanation for Statement I c. Statement I is True, Statement II is False . d. Statement I is false, Statement II is True. 1. A proton and an electron both have energy 50 eV. Statement I: Both have different wavelength. Statement II: Wavelength depends on energy and not on mass. |
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Answer» Correct Answer - C `lamda(h)/(sqrt(2mE))` |
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| 11. |
A laser used to weld detached retinas emits light with a wavelength of 652 nm in pulses that are 20.0ms in duration. The average power during each pulse is 0.6 W. then,A. the energy of each photon is `3.048xx10^(-19)J`B. the energy content in each pulse is 12mJC. the number of photons in each pulse is nearly `4xx10^(15)`D. the energy of each photon is nearly 1.9 eV |
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Answer» Correct Answer - A::B::C::D Energy of photon, `E_0=(hc)/(lamda)` `=(6.626xx10^(-34)xx3xx10^(8))/(652xx10^(-9))` `=3.048xx10^(-19)J=1.905eV` Energy content in each pulse is `0.6Wxx20xx10^(-3)s` `E_p=12xx10^(-3)J=12mJ` The number of photons in each pulse is `(E_p)/(E_0)=(12xx10^-3)/(3.048xx10^(-19))` `=3.9xx10^(16)approx4xx10^(16)` |
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| 12. |
A collimated beam of light of flux density `3kWm^(-2)` is incident normally on `100mm^(2)` completely absorbing screen. If P is the pressure exerted on the screen and `trianglep` is the momentum tranferred to the screen during a 1000 s interval, thenA. `P=10^(-3)Nm^(-2)`B. `P=10^(-4)Nm^(-2)`C. `trianglep=10^(-4)kgms^(-1)`D. `trianglep=10^(-5)kgms^(-1)` |
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Answer» Correct Answer - B::D Since `P=(I)/(c )=10^(4)Nm^(-2)` `P=(E)/(A)=(1)/(A)(trianglep)/(trianglet)` `trianglep=Patrianglet=10^(-5)kgms^(-1)` |
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| 13. |
A sensor is exposed for time t to a lamp of power P placed at a distance l. The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is `lamda` isA. `N=(Plamdad^2t)/(hcl^2)`B. `N=(4lamdad^2t)/(hcl^2)`C. `N=(Plamdad^2t)/(4hcl^2)`D. `N=(Plamdad^2t)/(16hcl^2)` |
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Answer» Correct Answer - A `E=(hc)/(lamda)` Number of photons emitted is `(Pt)/(((hc)/(lamda)))=n_0` `n_0(Plamdat)/(hc)` Since the radiation is spherically symmetric, so total number of photons entering the sensor is `n_0` times the ratio of aperture area to the area of a sphere of radius l. `N=n_0(pi(2d)^2)/(4pil^2)=(Plamdat)/(hc)(d^2)/(l^2)` |
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| 14. |
A pushed dye laser emits light of wavelength 585 nm. Because this wavelength is strongly absorbed by the haemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and geat of vaporization as water. [`S=4.2xx10^(3)J(kgK)^(-1)`,`L=2.25xx10^(6)Jkg`] Q. Suppose that each pulse must remove `2mug` of blood by evaporating it starting at `30^circ`C. The energy that each pulse must deliver to the blemish is nearlyA. `5.1J`B. `5.1mJ`C. `5.1muJ`D. `5.1kJ` |
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Answer» Correct Answer - B Energy required, `E=mstriangletheta+mL` `=2xx10^(-9)xx4.2xx10^(3)xx70+2xx10^(-9)xx2.25xx10^(6)` `=5.088xx10^(-3)J` `=3.18xx10^(16)eV` |
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| 15. |
A parellel beam of monochromatic light of wavelength 500 nm is incident normally on a perfectly absorbing surface. The power through any cross section of the beam is 10 W. Find (a) the number of photons absorbed per second by the surface and (b) the force exerted by the light beam on the surface. |
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Answer» The energy of each photon is `E=(hc)/(lamda)=((4.14xx10^(-15)eV-s)xx(3xx10^(8)(m)/(s)))/(500nm)` `=(1242eV-nm)/(500nm)=2.48eV` In 1 s, 10 J of energy passes through any cross section of the beam. Thus, the number of photons crossing a cross section per unit time is `n=(10J)/(2.48eV)=2.52xx10^(19)` This is also the number of photons falling on the surface per second and being absorbed. b. The linear momentum of each photon is `p=(h)/(lamda)=(hv)/(c )` The total momentum of all the photons falling per second on the surface is `(nhv)/(c)=(10J)/(c)=(10J)/(3xx10^(8)(m)/(s))=3.33xx10^(-8)N-s` As the photons are completely absorbed by the surface, this much momentum is transferred to the surface per second. The rate of change of the momentum of the surface, i.e., the force on it is `F=(dp)/(dt)=(3.33xx10^(-8))/(1s)=3.33xx10^(-8)N` |
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| 16. |
A cylindrical rod of some laser material `5xx10^-2`m long and `10^-2m` in diameter contains `2xx10^25` ions per `m^3`. If on excitation all the ions are in the upper energy level and de-excite simultaneously emitting photons in the same direction , calculate the maximum energy contained in a polse of radiation of wavlength `6.6xx10^-7`m. If the pulse lasts for `10^-7s`, calculate the average power of the laser during the pulse. |
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Answer» Total number of ions in the rod =number of ions per unit volume`xx` volume of rod `=(2xx10^(25))xx3.14xx(0.005)^(2)xx(5xx10^(-2))` `=7.85xx10^(19)` The number of photons excited in one direction is equal to the total number of ions because all ions are excited Noe, excited energy =number of excited photons`xx` energy of photon `(7.85xx10^(19))xx(hc)/(lamda)` `=(7.85xx10^(19))xx((6.6xx10^(-34))xx(3xx10^(8)))/(6.6xx10^(-7))` `=23.55W` |
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| 17. |
Calculate de Broglie wavelength for an average helium atom in a furnace at 400 K. Given mass of helium`=4.002`amu. |
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Answer» De Broglie wavelength, `lamda=(h)/(p)` Mean KE of helium atom `=(3kT)/(2)` or `E=(3xx(1.38xx10^(-23))xx400)/(2)` In terms of momentum, the energy is given by `E=(p^2)/(2m)` or `p^2=2mE` or `p=sqrt(2mE)` `p=sqrt(2xx(4.002xx1.66xx10^(-27))xx3xx(1.38xx10^(-23))xx200)` (where mass of helium`=4.002 a m u =4.002xx1.66xx10^(-27)kg`) Now, `lamda=(6.625xx10^(-34))/(p)` Substituting the value of p and solving it, we get `lamda=0.63xx10^(-10)`m |
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| 18. |
A sodium metal piece is illuminated with light of wavelength 0.3 `mum`. The work function of sodium is 2.46 eV. For this situation ,mark out the correct statement (s).A. The maximum kinetic energy of the ejected photoelectrons is 1.68 eVB. The cut-off wavelength for sodium is 505 nmC. The minimum photon energy of incident light for photoelectric effect to take place is 2.46 eVD. All of the above |
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Answer» Correct Answer - D `K_(max)=hv-phi` `=((6.63xx10^(-34)xx3xx10^(8))/(0.3xx10^(-6)xx1.6xx10^(-19))-2.46)eV` `=1.68eV` Cut-off wavelength, `lamda_0=(hc)/(phi)=505nm` The minimum energy required to eject the photo-electrons is equal to work function. |
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| 19. |
A particle of mass m is projected form ground with velocity u making angle `theta` with the vertical. The de Broglie wavelength of the particle at the highest point isA. `infty`B. `(h)/(m u sin theta)`C. `(h)/(m u costheta)`D. `(h)/(m u)` |
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Answer» Correct Answer - B Velocity at highest point`=usintheta` `lamda_D=(h)/(m usintheta)` (`sintheta` is angle w.r.t. verical) |
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| 20. |
A 100 W sodium lamp is radiating light of wavelength `5890A`, uniformly in all directions, a. At what rate, photons are emitted from the lamp? b. At what distance from the lamp, the average flux is 1 photon(`cm^2-s)^-1?` c. What are the photon flux and photon density at 2m from the lamp? |
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Answer» The energy of photon is given by `E=hupsilon=(hc)/(lamda)=(1990xx10^(-28)J)/(lamda)` `=(1990xx10^(-28))/(5890A)=3376xx10^(-22)J` ..(i) Given that the lamp is emitting energy at the rate of `100Js^(-1)` (power=100 W). Hence, number of photons N emitted is given by `N=(100)/(3376xx10^(-22))cong3xx10^(20)` photons `s^(-1)` ..(ii) b. We regard the lamp as a point source. Therefore, at a distance r from the lamp, the light energy is uniformly distributed over the surface of sphere of radius r. So, N photons are crossing area `4pir^2` of spherical surface per second. So, flux at a distance r is given by `n=(N)/(4pir^2)` or `r=sqrt(((N)/(4pi)))` or `r=sqrt(((3xx10^(20))/4xx3.14))cm=488860km` ...(iii) So, at this distance, on the average one photon will cross through `1 cm^(2)` area normal to tradial direction. c. Photon flux at `t=2m` `n=sqrt(((3xx10^(20))/(4xx3.14)))cm=5.9xx10^(14)` photons `cm^(-2)` Average density of photons at `r=2m `is given by `rho=(3xx10^(20))/(4xx3.14xx(200)^(2)xx(3xx10^(10)))` `=2xx10^(4)` photons `cm^(-2)` |
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| 21. |
A 60 W bulb is placed at a distance of 4 m from you. The bulb is emtting light of wavelength 600 nm uniformly in all directions. In 0.1 s, how many photons enter your eye if the pupil of the eye is having a diameter of 2mm? [take `hc=1240eV-nm]`A. `2.84xx10^(12)`B. `2.84xx10^(11)`C. `9.37xx10^(11)`D. `6.48xx10^(11)` |
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Answer» Correct Answer - B The intensity of light at the lacation of your eye is `I=(P)/(4pir^2)=(60)(4pixx4^2)Wm^(-2)` The energy entering into your eye per second is `P_1Ixx(pid^2)/(4)` where d is the diameter of pupil. `P_1=(60)/(4pixx4^2)xx(pixx(2xx10^(-3))^2)/(4)` `=9.375xx10^(-7) Js^(-1)` Let n be the number of photons enetering into the eye per second , then `P_1nxx(hc)/(lamda)` `9.375xx10^(-7)=nxx(1240xx1.6xx10^(-19))/(600)` `n=2.84xx10^(12)` photon `s^(-1)` So, the number of photons entering the eye in `0.1 s=0.1n=2.84xx10^(11)` |
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| 22. |
What amount of energy sould be added to an electron to reduce its de Broglie wavelength `lamda_1=550nm` incident on it, causing the ejection of photoelectrons for which the stopping potential is `V_(s1)=0.19V`. If the radiation of wavelength `lamda_2=190nm` is now incident on the surface, (a) calculate the stopping potential `V_(S2)`, (b) the work function of the surface, and (c) the threshold frequency for the surface. |
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Answer» We know that, de Broglie wavelength `lamda=(h)/(mv)` and `R=(1)/(2)mv^2` `lamda=(h)/(sqrt(2mE))` In first case, `100xx10^(-12)=(h)/(sqrt(2mE_(1))` …(i) In second case, `50xx10^(-12)=(h)/(sqrt(2mE_(2))` ..(ii) Dividing Eqs. (i) by (ii), we get `2=sqrt(((E_2)/(E_1)))` or `E_2=4E_1` So, energy to be added `=4E_1-E_1=3E_1` Now, `(h)/(sqrt(2mE_1))=100xx10^(-12)` or `sqrt(2mE_1)=(6.625xx10^(-34))/(10^(-10))` or `sqrt(2mE_1)=6.625xx10^(-24)` or `E_1=((6.625xx10^(-24))^(2))/(2xx(9.1xx10^(-31)))=150eV` There, energy added `=3E_1=450eV` |
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| 23. |
A surface has light of wavelength `lamda_1=550nm` incident on it, causeing the ejection of photoelectrons for which the shopping potential is `V_(s1)=0.19V`. If the radiation of wavelength `lamda_2=190nm` is now incident on the surface, (a) calculate the stopping potential `V_(s2)`, (b) the work function of the surface, and (c) the threshold frequency for the surface. |
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Answer» For the first wavelength : `eV_(s1)=hupsilon-W` ...(i) `W=`work function: `eV_(s1)=hupsilon_2-W` ...(ii) Subtracting, we get `V_(S2)-V_(S1)=(h)/(e)(upsilon_2-upsilon_1)` or `V_(S2)=V_(S1)+(hc)/(e)((1)/(lamda_2)-(1)/(lamda_1))` `V_(S1)+(hc)/(e)((lamda_1-lamda_2)/(lamda_1lamda_2))` `=0.19+1240((550-190)/(190xx550))=4.47V` b. From Eq. (i): `W=(hc)/(lamda_1)-eV_(S1)` Work function (in eV) `(W)/(e)=(hc)/(elamda_1)-V_(S1)=(1240)/(550)-0.19=2.07eV` c. Threshold frequency: `V_0=(W)/(h)=(2.07xx1.6xx10^(-19))/(6.62xx10^(-34))=5xx10^(14)Hz` |
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| 24. |
in Q-82, the work function isA. 0.212 eVB. 0.313 eVC. 0.414 eVD. 0.515 eV |
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Answer» Correct Answer - C `phi_0=hc_0` or `phi_0=exx4.14xx10^(-15)xx1xx10^(14)` `=0.414 eV` |
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| 25. |
Two identical metal plates show photoelectric effect by a light of wavelength `lambda_(A)` falls on plate `A` and `lambda_(B)` on plate `B(lambda_(A) = 2 lambda_(B))`. The maximum kinetic energy isA. `2K_(A) = K_(B)`B. `K_(A) lt K_(B)//2`C. `K_(A) = 2K_(B)`D. `K_(A) gt K_(B)//2` |
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Answer» Correct Answer - B `(hc)/(lambda_(A)) =phi+K_(A)` ...(i) `(hc)/(lambda_(B)) =phi+K_(B)` …(ii) Let `lambda_(A) = lambda, lambda_(B) =lambda//2` `(K_(B))/(K_(A)) =((hc)/(lambda//2)-phi)/((hc)/(lambda)-phi) =((hc)/(lambda)-phi)/((hc)/(lambda)-phi) +((hc)/(lambda))/((hc)/(lambda)-phi)` `=1 +(gt1) = gt 2` `K_(B)gt2 K_(A)` `K_(A)lt(K_(B))/(2)` |
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| 26. |
Two identical metal plates show photoelectric effect. Light of wavelength `lamda_A` falls on plate A and `lamda_B` fall on plate B `lama_A=2lamda_B`, The maximum KE of the photoelectrons are `K_A` and `K_B`, respectively, Which one of the following is true?A. `2K_A=K_B`B. `K_A=2K_B`C. `(K_AltK_B/(2))`D. `K_Agt2K_B` |
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Answer» Correct Answer - C `K_A=(hc)/(lamda_A)-phi_0`,`K_B=(hc)/(lamda_B)-phi_0` But `lamda_A=2lamda_B`, therefore `K_A=(hc)/(2lamda_B)-phi_0` `K_A=(1)/(2)[K_B+phi_0]-phi_0` or `K_A=(K_B)/(2)-(phi_0)/(2)` `K_Alt(K_B)/(2)` |
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| 27. |
The potential difference applied to an X-ray tube is V The ratio of the de Broglie wavelength of electron to the minimum wavlength of X-ray is directrly proportional toA. V voltB. `sqrtV`C. `V^((3)/(2))`D. `V((7)/(2))` |
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Answer» Correct Answer - B `lamda=(h)/(sqrt(2meV))` `(hc)/(lamda_(min))=eV` `lamdaxx(hc)/(lamda_(min))=(h)/(sqrt(2meV))eV` or `(lamda)/(lamda_(min))propsqrtV` |
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| 28. |
The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let `lambda_1`be the de-Broglie wavelength of the proton and `lambda_2` be the wavelength of the photon. The ratio `(lambda_1)/(lambda_2)` is proportional to (a)`E^0` (b) `E^(1//2)` (c ) `E^(-1)` (d)`E^(-2)`A. `E^0`B. `E^(1)/(2)`C. `E^-1`D. `E^-2` |
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Answer» Correct Answer - B `lamda_1=(h)/(p)=(h)/(sqrt(2mE))` `lamda_2=(hc)/(E)=E^((1)/(2))` |
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| 29. |
With respect to electromegnetic theory of light, the photoelectric effect is best explained by statementA. Light waves carry energy and when light is incident on the metallic surface, the energy absorbed by the metal may somehow concentrate on individual electrons and reappear as their kinetic energy when ejectedB. Particles of light (photons) collide with the metal and the electrons take this energy and may ejectC. When light waves fall on a metallic surface, the stability of atoms is disturbed and the electrons come out to make the system stableD. none of the above |
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Answer» Correct Answer - A Option (a) correctly explains the photoelectric effect on the basis of electromagnetic theory. Its correct explanation is given by Quantum theory of light. |
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| 30. |
Explain Briefly how classical theory could not explain the phenomenon of photoelectric effect |
| Answer» To eject an electron from a metal surface, a minimum amount of energy, called the work function of the metal, is required Classically, if incident radiation is a wave its energy is shared by all the atoms in the metal surface and calculation show that it will take almost a year, when the requisite amount of energy may be accumulated by an electron. Hence , on classical picture, The photoelectric emission cannot be instantaneous. When the incident radiation is assumed to be sonsisting of photons, a photon of frequency `v_0` or more can possess the energy needed to eject an electron. If frequency of hte incident photon is less than `v_0`, no photoelectric emission will take place. In a way, `v_0` is cut-off frequency. | |
| 31. |
If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (Mass of the helium nucleus is `4.0026` amu and mass of oxygen nucleus is `15.9994` amu)A. 7.6 MeVB. 56.12 MeVC. 10.24 MeVD. 23.9 MeV |
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Answer» Correct Answer - C `4_2^4Herarr` `8^16O` `B.E=trianglemxx931.5MeV` `=(4xx4,0026-15.9994)xx931.5` `=10.24MeV` |
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| 32. |
A point source of light is used in a photoelectric effect. If the source is removed farther from the emitted metal, the stopping potentialA. will increaseB. will decreaseC. sameD. none |
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Answer» Correct Answer - C Stopping potential is independent of intensity |
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| 33. |
If the frequency of light in a photoelectric experiment is doubled the stopping potential willA. be doubledB. be halvedC. gt doubleD. lt double |
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Answer» Correct Answer - C `h upsilon = phi + eV_(s_(1))` …(i) `h.2 upsilon =phi + eV_(s_(2))` …(ii) `(V_(s_(2)))/(V_(s_(1))) =(2h upsilon -phi)/(h upsilon -phi) =(h upsilon -phi)/(h upsilon -phi) +(h upsilon)/(h upsilon -phi)` `=1+(gt1) = gt2` `V_(s_(2)) gt 2V_(s_(1))` |
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| 34. |
If the wavelength of light in an experiment on photoelectric effect is doubled, (i) the photoelectric emission will not take place (ii) the photoelectric emission may or may not take place (iii) the stopping potential will increase (iv) the stopping potential will decreaseA. (i),(ii)B. (ii),(iii)C. (i),(iv)D. (ii),(iv) |
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Answer» Correct Answer - D It mat be that `(hc)/(2 lambda) gt phi` or `(hc)/(2 lambda) lt phi` `V_(s)` decreases if wavelength increases |
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| 35. |
Light of frequeny 1.5 times the threshold frequency , imcident on a photo sensitive material .If the frequency of incident light is halved and the intensity is doubled the photo current becomesA. four timesB. doubleC. halfD. zero |
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Answer» Correct Answer - D Incident frequency `upsilon =(1.5upsilon_(0))/(2) =0.75upsilon_(0)` , since `upsilon lt upsilon_(0)`, no photoelectric effect |
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| 36. |
How many photons are emitted per second by a 5 mW laser source operating at 632.8 nm?A. `1.6xx10^(16)`B. `1.6xx10^(13)`C. `1.6xx10^(10)`D. `1.6xx10^(3)` |
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Answer» Correct Answer - A `E=(hc)/(lamda)=(6.6xx10^(-34)xx3xx10^(8))/(632.8xx10^(-9))J` Number of photons per second `n=(P)/(E)` `impliesn=(5xx10^(-3))/(3.14xx10^(-19))=1.6xx10^(16)` |
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| 37. |
The photoelectric threshold for some material is 200 nm. The material is irradiated with radiations of wavelength 40 nm. The maximum kinetic energy of the emitted photoelectrons isA. 2 eVB. 1 eVC. 0.5 eVD. none of these |
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Answer» Correct Answer - D Work function `W_0=(hc)/(lamda_0)` `implies=W_0=(2xx10^(-25))/(2xx10^(-7))=10^(-18)J=(10^(-18))/(1.6xx10^(-19))eV=6.25eV` energy of incident ratiation is `E=(hc)/(lamda)=31.25eV` KE of photoelectrons `=E-W_0=25eV` |
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| 38. |
Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,A. bot p and E increaseB. p increases E decreasesC. p dereases and E increasesD. both p and E decreases |
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Answer» Correct Answer - A `p=(h)/(lamda)` also, `E=(hc)/(lamda)` So, if `lamda` is decreased, both p and E increase. |
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| 39. |
The KE of the photoelectrons is E when the incident wavelength is `(lamda)/(2)`. The KE becomes 2E when the incident wavelength is `(lamda)/(3)`. The work function of the metal isA. `(hc)/lamda`B. `(2hc)/(lamda)`C. `(3hc)/(lamda)`D. `(hc)/(3lamda)` |
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Answer» Correct Answer - A `E=(hc)/((lamda)/(2))-phi_0` or `2E=(hc)/((lamda)/(3))-phi_0` or `2((2hc)/(lamda)-phi)=(3hc)/(lamda)-phi_0` or `(4hc)/(lamda)-2phi_0=(3hc)/(lamda)-phi_0` or `phi_0=(hc)/(lamda)` |
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| 40. |
In a photoelectric effect experimetent, a metallic surface of work function 2.2 eV is illuminated with a light of wavelenght 400 nm. Assume that an electron makes two collisions before being emitted and in each collision 10% additional energy is lost. Q. After how many collisions will the electron be unable to come out of the metal?A. 2B. 6C. 4D. 8 |
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Answer» Correct Answer - C Energy of photon `E=(hc)/(lamda)=(1.24xx10^(3))/(400)=3.1 eV` Remaining energy `=3.1-0.31=2.79eV` Energy lost is first collision is `(3.1)xx((10)/(100))=0.31eV` Remaining energy is `3.1-0.31=2.79 eV` Energy lost in second collision is `(2.79)xx((10)/(100))=0.279 eV` Total energy lost two collisions is `(0.31)+(0.279)eV=0.589 eV` So, from conservation of energy, we have `(hc)/(lamda)=phi+KE_(max)+` energy lost in two collision `3.1=2.2+KE_(max)+0.589` `KE_(max)=0.31eV` Total energy after second collision is `(2.79-0.279)=2.511 eV` Energy lost in third collision is `2.511xx(10)/(100)=0.2511 eV` Remaining energy `=(2.511-0.2155)=2.2500 eV` Energy lost in fourth collision `=(2.2599xx(10)/(100))=0.2259 eV` Remaining energy `=(2.2599-0.2250)=2.034 eV` After the fourth collision, the electron doen not have enough energy to overcome the work function, so it cannot come out. |
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| 41. |
When a centimeter thick surface is illuminated with light of wavelength `lamda`, the stopping potential is V. When the same surface is illuminated by light of wavelength `2lamda`, the stopping potential is `(V)/(3)`. Threshold wavelength for the metallic surface isA. `(4lamda)/(3)`B. `4lamda`C. `6lamda`D. `(8lamda)/(3)` |
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Answer» Correct Answer - B `eV=hc((1)/(lamda)-(1)/(lamda_0))` `(eV)/(3)=hc((1)/(2lamda)-(1)/(lamda_0))` Dividing Eq. (i) and (ii), we get `lamda_0=4lamda` |
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| 42. |
When a certain metallic surface is illuminated with monochromatic light of wavelength `lamda`, the stopping potential for photoelectric current is `3V_0` and when the same surface is illuminated with light of wavelength `2lamda`, the stopping potential is `V_0`. The threshold wavelength of this surface for photoelectrice effect isA. `6 lambda`B. `3 lambda`C. `4 lambda`D. `8 lambda` |
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Answer» Correct Answer - C `(hc)/(lambda) =phi +e.3 V_(0)` ….(i) `(hc)/(2lambda) =phi + e.V_(0)` ….(ii) 3(ii) - (i) `implies (hc)/(2 lambda) =3 phi - phi =2 phi =2. (hc)/(lambda_(0))` `lambda_(0) =4 lambda` |
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| 43. |
Photons of energy 5eV are incident on the cathode. Electrons reaching the anode have kinetic energies varying from 6eV to 8eV. Find the work function of the metal and state whether the current in the circuit is less than or equal to saturation current. |
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Answer» `KE_(max)=(5-phi)eV` When electrons are accelerated through 5V, they will reach the anode with energy `=(5-phi+5)eV` `10-phi=8` or `phi=2eV` Current is less than saturation current. This is because if the slowest electron also reached the plate it would have 5 eV energy at the anode, but it is given that the minimum energy is 6 eV. |
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| 44. |
If a suface has work function of 3.00eV, the longest wavelength of light which will cause the emission of electrons isA. `4.8xx10^-7`mB. `5.99xx10^-7`mC. `4.13xx10^-7`mD. `6.84xx10^-7`m |
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Answer» Correct Answer - C `(hc)/(lamda_(max))=3xx1.6xx10^-19J` `implieslamda_(max)=(6.6xx10^-34xx3xx10^8)/(3xx1.6xx10^-19)` `=4.125xx10^-7`m |
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| 45. |
The human eye is most sensitive to green light of wavelength 505 nm. Experiments have found tthat when people are kept in a dark room until their eye adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina. The velocity of typical bacterium of mass `9.5xx10^(-12)`g, if it had absorbed all energy of photon ,is nearlyA. `10^(-6)ms^(-1)`B. `10^(-8)ms^(-1)`C. `10^(-10)ms^(-1)`D. `10^(-13)ms^(-1)` |
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Answer» Correct Answer - D Momentum of photon of green light, `p=(h)/(lamda)=(6.626xx10^(-34))/(505xx10^(-9))1.3xx10^(-27)kgms^(-1)` Velocity of bacterium, `v=(p)/(m)=(1.3xx10^(-27))/(9.5xx10^(-15))` `=1.368xx10^(-13)ms^(-1)` |
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| 46. |
Radiation of wavelength 546 nm falls on a photo cathode and electrons with maximum kinetic energy of 0.18 eV are emitted. When radiation of wavelength 185 nm falls on the same surface, a (negative) stopping potential of 4.6 V has to be applied to the collector cathode to reduce the photoelectric current to zero. Then, the ratio `(h)/(e)` isA. `6.6xx10^(-15)JsC^(-1)`B. `4.14xx10^(-15)JsC^(-1)`C. `6.6xx10^(-34)JsC^(-1)`D. `4.12xx10^(-34)JsC^(-1)` |
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Answer» Correct Answer - B `h(v_1-v_2)=e(V_1-V_2)` `(h)/(e)=(1)/(c)((V_1-V_2))/(((1)/(lamda_1)-(1)/(lamda_2)))=(10^(-9))/(3xx10^(8))((4.6-0.08)/((1)/(185)-(1)/(546)))` `=(4.42xx185xx546xx10^(-17))/(361xx31)` `=4.12xx10^(-15)JsC^(-1)` |
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| 47. |
Lights of two different frequencies whose photons have energies 1 and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electronsA. `1:5`B. `1:4`C. `1:2`D. `1:1` |
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Answer» Correct Answer - C If E is the energy of incident photon and W the work function, then `E_W_0="available energy"` `E-W_0=(1)/(2)mv^(2)` or `v=sqrt((2(E-W_0))/(m))` `(v_1)/(v_2)=sqrt((1-0.5)/(2.5-0.5))=sqrt((0.5)/(2))=(1)/(2)` |
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| 48. |
A proton when accelerated through a potential difference of V volt has a wavelength `lamda` associated with it. An alpha-particle in order to have the same `lamda` must be accelerated through a potential difference ofA. V voltB. 4V voldC. 2 V voltD. `((V)/(8))` volt |
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Answer» Correct Answer - D `lamda_P=lamda_alpha` or `(h)/(sqrt(2m_pQ_pV))=(h)/(sqrt(2m_alphaQ_alphaV_alpha))` `m__pQ_pV_p=m_alphaQ_alphaV_alpha` `V_alpha=((m_p)/(m_alpha))((Q_p)/(Q_alpha))V=((1)/(4))((1)/(2))V=(V)/(8)` |
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| 49. |
The de Broglie wavelength of neutrons in thermal equilibrium is (given `m_n=1.6xx10^(-27)`kg)`A. `(30.8)/(sqrtT)A`B. `(3.08)/(sqrtT)A`C. `(0.308)/(sqrtT)A`D. `(0.0308)/(sqrtT)A` |
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Answer» Correct Answer - A `lamda=(h)/(sqrt(2meT))` `=(6.62xx10^(-34))/(sqrt(2xx1.67xx10^(-27)xx1.38xx10^(-23)T))m` `=(6.62xx10^(-34))/(2.15xx10^(-25)sqrtT)m` `(3.079)/(sqrtT)xx10^(-9)m=(30.79)/(sqrtT)Aapprox(30.8)/(sqrtT)A` |
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| 50. |
A partical of mass M at rest decays into two Particles of masses `m_1 and m_2` having non-zero velocities. The ratio of the de - Broglie wavelengths of the particles `lambda_1| lambda_2` is (a) `m_1//m_2` (b)`m_2// m_1` (c ) 1 (d)`sqrt2 // sqrt1`A. `(m_2)/(m_1)`B. `(m_1)/(m_2)`C. `(sqrtm_1)/(sqrtm_2)`D. `1:1` |
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Answer» Correct Answer - D `lamda=(h)/(mv)` Here, `0xxM=m_1v_1+m_2v_2` Clearly, `m_1v_1=-m_2v_2` In magnitude, mv `=` constant `(lamda_1)/(lamda_2)=(1)/(1)` |
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