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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. |
When a metal surface is illuminated by a monochromatic light of wave-length `lambda`, then the potential difference required to stop the ejection of electrons is `3V`. When the same surface is illuminated by the light of wavelength `2lambda`, then the potential difference required to stop the ejection of electrons is `V`. Then for photoelectric effect, the threshold wavelength for the metal surface will beA. `6lambda`B. `4lambda//3`C. `4lambda`D. `8lambda` |
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Answer» Correct Answer - C `(hc)/lambda=(hc)/lambda_(0)+eV_(0)` or `hc(1/lambda-1/lambda_(0))=eV_(0)` |
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| 2. |
Which of the following is dependent on the intensity of incident radiation in a photoelectric experimentA. work function of the surfaceB. amount of photoelectric currentC. stopping potentialD. maximum kinetic energy |
| Answer» Correct Answer - B | |
| 3. |
A radiation of wavelength 200nm is propagating in the form of a parallel surface. The intensity of the beam is 5mW and its cross-sectional area is 1.0`mm^(2)`. Find the pressure exerted by radiation on the metallic surface if the radiation is completely reflected. |
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Answer» `:.E=12400/lambda=12400/200=6.2eV~~10^(-18)J` Number of photons passing a point per second is `n=P/E=(5xx10^(-9))/10^(-18)=5xx10^(9)`.momentum of each photon `p=E/C=3.3xx10^(-27)J//s`. Change in momentum after each stick `=2p=6.6xx10^(-27)J//s` Total momentum change per second is `F=(dp)/(dt)=(nxx2p)/t=5xx10^(9)xx6.6xx10^(-27)=33xx10^(-18)N` `:.pressure F/A=33xx10^(-12)N/m^(2)` |
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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 `(i)` and `(ii)` are trueB. `(i)` is true but `(ii)` is falseC. `(i)` is false but `(ii)` is trueD. Both `(i)` and `(ii)` are false |
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Answer» Correct Answer - B `I=(nhv)/(tA)`, when frequency is doubled for `I` to become doubled `n` should remain the same. `therefore` saturation photo current remains the same but maximum kinetic energy becomes more than doubled (from `KE_(max)=hv-w`) |
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| 5. |
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. `7.5xx10^(15) eV,2.7 eV`B. `6.5xx10^(16) eV,2.9 eV`C. `6.5xx10^(16) eV,2.7 eV`D. `7.5xx10^(16) eV,1.9 eV` |
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Answer» Correct Answer - D `E=pt,E("in each photon" )=(hc)/lambda` |
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| 6. |
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. `oo`B. `h/("mu" sin theta)`C. `h/("mu" cos theta)`D. `h/"mu" ` |
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Answer» Correct Answer - B `lambda=h/(mv)=h/(mu sin theta)` |
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| 7. |
Two initially uncharged concentric than conducting spherical shells of radius `a` and `2a` are as shown and the inner shell is grounded. The work function of outer shell is `phi_(0)`.At time `t=0`, a continuous parallel beam of monochromatic radiation of cross-section area `A` and intensity `I` is incident on outer shell.The energy of each photon is `hv` such that `hv gt phi_(0)` Assuming for each incident photon one photoelectron is ejected, answer the following questions. The time after `t=0`, at which charge on outer sphere becomes constant. A. `(hupsilon-phi_(0))/e`B. `(hupsilon-phi_(0))/(2e)`C. `2/3(hupsilon-phi_(0))/e`D. None of these |
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Answer» Correct Answer - A The potential difference is `Vs=(hv-phi)/e` |
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| 8. |
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-broglie wavelength of the particles varies cyclically between two values `lambda_(1), lambda_(2) with lambda_(1)gtlambda_(2)`. Which of the following statements are true?A. The particle could be moving in a circular orbit with origin as centreB. The particle could be moving in an elliptic orbit with origin as its focusC. When the de Broglie wave length is `lambda_(1)` the particle is nearer the origin that when its value is `lambda_(2)`D. When the de Broglie wavelength is `lambda_(2)` the particle is nearer the origin that when its value is `lambda_(1)` |
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Answer» Correct Answer - B::D Angular momentum `(mvr)` of a particle is conserved, when it moves in an elliptical orbit.As `r` changes for elliptical orbit,`mv` also changes and so does `lambda(=h//mv)` When particle is nearer to the origin,`r` is less,`mv` is more and hence `lambda` is less. `lambda_(1)=h/(mv_(1))` and `lambda_(2)=h/(mv_(2))` `therefore lambda_(1)/lambda_(2)=v_(2)/v_(1)`,since `lambda_(1) gt lambda_(2) therefore v_(2) gt v_(1)` |
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| 9. |
The maximum K.E. of photoelectrons ejected from a photometer when it is irradiated with radiation of a wavelength 400nm is 1eV. If the threshold energy of the surface is 1.9 eV,A. The maximum `K.E`. Of photo electrons when it is irradiated with `500nm` photons will be `0.42 eV`B. The maximum `K.E`. In case `(a)` will be `1.725eV`C. The longest wavelength which will eject the photo electrons from the surface is nearly `610nm`D. The maximum `K.E`. Will increase if the intesity of radiation is increased |
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Answer» Correct Answer - A::C `KE_(max)=(hc)/(elambda)-WrArr(hc)/(elambda)=1+1.9=2.9eV` `rArr (hc)/e=2.9xx400 eV nm` for `lambda=500 nm,E=(hc)/(e500)=(2.9xx4cancel(0)cancel(0))/(5cancel(0)cancel(0))=2.32eV` i.e `KE_(max)=2.32-1.9=0.42eV` `therefore lambda_(0)=(hc)/(eW)=(2.9xx400)/1.9=610 A^(0)` |
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| 10. |
In photoelectric effect, which of the following property of incident light will not affect the stopping potentialA. FrequencyB. WavelengthC. EnergyD. Intensity |
| Answer» Correct Answer - D | |
| 11. |
A non-monochromatic light is used in an experiment on photoelectric effect. The stopping potentialA. is related to the mean wavelengthB. is related to the longest wavelengthC. is related to the shortest wavelengthD. is not related to the wavelength |
| Answer» Correct Answer - C | |
| 12. |
Two separate monochromatic light beams `A` and `B` of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength are `lambda_(A)` and `lambda_(B)` respectively. Assuming that all the the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam `A` to that from `B` isA. `(lambda_(A)/lambda_(B))`B. `(lambda_(B)/lambda_(A))`C. `(lambda_(A)/lambda_(B))^(2)`D. `(lambda_(B)/lambda_(A))^(2)` |
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Answer» Correct Answer - A The number of photo electron depends on the Number of photons Number of photon `=I/(hc//lambda)=(lambda.I)/(hc) prop lambda` Ratio of no. of photo electrons =`lambda_(A)/lambda_(B)` |
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| 13. |
A photocell is illuminated by a small bright source places `1` m away when the same source of light is placed `(1)/(2)` m away. The number of electron emitted by photocathode would beA. increase by factor of `2`B. decrease by a factor of `2`C. increase by a factor of `4`D. decrease by a factor of `4` |
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Answer» Correct Answer - C `Ialpha1/r^(2)` |
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| 14. |
When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then (a) the stopping potential will be 0.2 V (b) the stopping potential will be 0.6 V (c ) the saturation current will be 6.0 mA (d) the saturation current will be 2.0 mAA. the stopping potential will be `0.2V`B. the stopping potential will be `0.6V`C. the stopping potential will be `6.0V`D. the stopping potential will be `2.0V` |
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Answer» Correct Answer - B::D (b)Stopping potentail remains the same as it depends on the frequency of incident radiation. (D)Saturation current `alpha` intensity of incident radiation `a1/r^(2)`.Since `r` becomes three times `((0.6m)/(0.2m))`,saturation current becomes `(18.0mA)/(3)^(2)=2.0mA` |
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| 15. |
In Davisson-Germer electron diffraction arrangement if suppose the voltage applied to accelerated electrons is increased, the value of the angles at which diffracted beam have the maximum intensityA. will be larger than the earlier value.B. will be the same as the earlier value.C. will be less than the earlier value.D. will depend on the target. |
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Answer» Correct Answer - C As `lambda=12.27/sqrtVoverset(0)A`, when potential `(V)` is increased,`lambda`decreases as `2d sin theta=n lambda, theta` will be less than the earlier value. |
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| 16. |
If an electron and a proton have the same `KE`, the ratio of the de Broglie wavelengths of proton and electron would approximately beA. `1:1837`B. `43:1`C. `1837:1`D. `1:43` |
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Answer» Correct Answer - D `lambda=h/sqrt(2mE)` |
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| 17. |
From Davisson-Germer experiment an `alpha` particle and a proton are accelerated through the same `pd V`. Find the ratio of the de Broglie wavelengths associated with themA. `1:2sqrt2`B. `2sqrt2:1`C. `1:sqrt2`D. `sqrt2:1` |
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Answer» Correct Answer - A `lambda=h/p=h/sqrt(2mq V)` |
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| 18. |
A proton and an `alpha`-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths `lambda_(p) and lambda_(a)` related to each other?A. `sqrt2`B. `sqrt3`C. `sqrt8`D. `sqrt10` |
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Answer» Correct Answer - C As,`lambda=h/sqrt(2mqv),thereforelambdaprop1/sqrt(mq)` `lambda_(p)/lambda_(alpha)=sqrt(m_(alpha)q_(alpha))/sqrt(m_(p)q_(q))=sqrt(4m_(p)xx2e)/sqrt(m_(p)xxe)=sqrt8` `therefore lambda_(p)=sqrt8lambda_(alpha)` i.e., wavelength of proton is `sqrt8` times wavelength of `alpha`-particle. |
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| 19. |
A potential of 10000 V is applied across an x-ray tube. Find the ratio of de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays. |
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Answer» Debroglie wave length of incident electron is `lambda_(1)=h/sqrt(2meV)`...1 Shortest wavelength of `x`ray photon is `lambda_(2)=(he)/(Ve)`...(2) `rArr lambda_(1)/lambda_(2)=1/csqrt((V/2)(e/m))=0.1` |
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| 20. |
A proton, a neutron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare asA. `lambda_(p)=lambda_(n)gtlambda_(e)gtlambda_(alpha)`B. `lambda_(alpha) lt lambda_(p)=lambda_(n)gtlambda_(e)`C. `lambda_(e)ltlambda_(p)=lambda_(n)gtlambda_(alpha)`D. `lambda_(e)=lambda_(p)=lambda_(n)=lambda_(alpha)` |
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Answer» Correct Answer - B We know that the relation between `lambda` and `K` is given by `lambda=h/sqrt(2mk)` Here, for the given value of energy `K,h/sqrt(2k)` is a constant. Thus, `lambda prop 1sqrtm` `therefore lambda_(p):lambda_(n):lambda_(e):lambda_(alpha)` `1/sqrtm_(p):1/sqrtm_(n):1/sqrtm_(e):1/sqrtm_(alpha)` Since, `m_(p)=m_(n)`, hence `lambda_(p)=lambda_(alpha)` As,`m_(alpha) gt m_(p)`, therefore `lambda_(alpha) lt lambda_(p)` As,`m_(e) lt m_(n)`, therefore `lambda_(e) gt lambda_(n)` Hence,`lambda_(alpha) lt lambda_(p)=lambda_(n) lt lambda_(e)` |
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| 21. |
The ratio of de - Broglie wavelength of `alpha `- particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector `vec(B)` is perpendicular to the velocity vectors of the `alpha` - particle and the proton is |
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Answer» Correct Answer - B `|barF|=|(-dU)/(dr)|=momega^(2)r , (mv^(2))/r=momega^(2)r , V=omegar`..(i) `mvr=(nh)/(2pi),V=(nh)/(2pimr)`...(ii) (i) & (ii) `rArr r=[(nh)/(2pimomega)]^(1/2) rArr r prop n^(1/2)` `therefore x=2` |
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| 22. |
The wavelength of de-Broglie wave associated with a thermal neutron of mass `m` at absolute temperture `T` is given by (here, `k` is the Boltzmann constant)A. `h/sqrt(2mKT)`B. `h/sqrt(mkT)`C. `h/sqrt(3mkT)`D. `h/(2sqrt(3mkT))` |
| Answer» Correct Answer - C | |
| 23. |
The magnitude of the de-Broglie wavelength `(lambda)` of an electron `(e)`,proton`(p)`,neutron `(n)` and `alpha` particle `(a)` all having the same energy of `Mev`, in the incresing order will follow the sequence:A. `lambda_(e),lambda_(p),lambda_(n),lambda_(alpha)`B. `lambda_(alpha),lambda_(n),lambda_(p),lambda_(e)`C. `lambda_(e),lambda_(n),lambda_(p),lambda_(alpha)`D. `lambda_(p),lambda_(e),lambda_(alpha),lambda_(n)` |
| Answer» Correct Answer - B | |
| 24. |
The wavelength of a proton and a photon are same. ThenA. Their velocities are sameB. Their momenta are equalC. Their energies are sameD. Their speeds are same |
| Answer» Correct Answer - B | |
| 25. |
The momentum of a proton is `p`. the corresponding wavelength isA. `h//p`B. `h p`C. `p//h`D. `sqrt(hp)` |
| Answer» Correct Answer - A | |
| 26. |
In a photoelectric effect experiment, photons of energy `5eV` are incident on a metal surface They liberate photoelectron which are just stopped by an electrode at a potential of `-3.5V` with respect to the metal. The work fuction of the metal isA. `1.5eV`B. `3.5eV`C. `5.0eV`D. `8.5eV` |
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Answer» Correct Answer - A `E=w+eV_(0)` |
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| 27. |
Though quantum theory of light can explain a number of phenomena observed with light , it is necessary to retain the wave-nature of light to explain the phenomena of :A. photoelectric effectB. diffractionC. compton effectD. black body radiation |
| Answer» Correct Answer - B | |
| 28. |
In each of the following questions, a statement is given and a corresponding statement or reason is given just below it. In the statement, mark the correct answer as Statement-I: Though light of a single frequency (monochromic light) is incident ona metal, the energies of emitted photoelectrons are different Statement-II: The energy of electrons just after they absorb photons incident on the metal surface may be lost in collision with other atoms in the metal before the electron is ejected out of the metal.A. Statement `I` is true,Statemetn `II` is true, statement `II` is a correct explanation of statement `I`.B. Statement `I` is true, Statement `II` is true,Statement `II` is `NOT` a correct explanation for statement `I`.C. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true |
| Answer» Correct Answer - A | |
| 29. |
Relativistic corrections become necessary when the expression for the kinetic energy `1/2mv^(2)`, becomes comparable with `mc^(2)`, where m is the mass of the particle. At what de-broglie wavelength will relativistic corrections become important for an electron?A. `lambda=10 nm`B. `lambda=10^(-1) nm`C. `lambda=10^(-4) nm`D. `lambda=10^(-6) nm` |
| Answer» Correct Answer - C::D | |
| 30. |
A proton when accelerated through a `p.d` of `V` volt has wavelength `lambda` associated with it. An electron to have the same `lambda` must be accelerated through a `p.d` ofA. `V/8` voltB. `4V` voltC. `2V` voltD. `1838V` volt |
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Answer» Correct Answer - D `V_(1)xxq_(1)xxm_(1)=V_(2)xxq_(2)xxm_(2)` |
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| 31. |
If the momentum of an electron is changed by `p`, then the de - Broglie wavelength associated with it changes by `0.5 %`. The initial momentum of electron will beA. `p_(m)//200`B. `p_(m)//100`C. `200p_(m)`D. `100p_(m)` |
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Answer» Correct Answer - C `(Deltalambda)/lambda=(-DeltaP)/P` |
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| 32. |
The de Broglie wavelength associated with an electron of velocity `0.3c` and rest mass `9.1xx10^(-31) kg` isA. `7.68xx10^(-10) m`B. `7.68xx10^(-12) m`C. `5.7xx10^(-12) m`D. `9.1xx10^(-12) m` |
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Answer» Correct Answer - B `lambda=(hsqrt(1-v^(2)/c^(2)))/(m_(0)V)` |
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| 33. |
The particles that can be accelarted by an electric field isA. protonB. electronC. alpha particleD. all above |
| Answer» Correct Answer - D | |
| 34. |
An electron of mass `9.1xx10^(-31) kg` and charge `1.6xx10^(-19) C` is accelarted through a potential difference of `V` volt. The de Broglie wavelength `(lambda)` associated with the electron isA. `12.27/sqrtV A^(0)`B. `12.27/V A^(0)`C. `12.27sqrtV A^(0)`D. `1/(12.27sqrtV) A^(0)` |
| Answer» Correct Answer - A | |
| 35. |
A particle of mass `M` at rest dacays into two particle of masses `m_(1)` and `m_(2) `having non zero velocity . The ratio of the de Broglie wavelength . The ratio of the de Broglie wavelength of the particle `lambda , _(1) // lambda_(2)` isA. `m_(1)/m_(2)`B. `m_(2)/m_(1)`C. `1:1`D. `sqrt(m_(2)/m_(1))` |
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Answer» Correct Answer - C `lambda=h/p` and from conservation of linear momentum both particles will have same magnitude of linear momentum after decay. |
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| 36. |
If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will beA. `41%`B. `141%`C. `100%`D. `71%` |
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Answer» Correct Answer - C `lambda alpha 1/sqrtE` |
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| 37. |
If the velocity of a particle is increased three times, then the percentage decrease in its de Broglie wavelength will beA. `33.3%`B. `66.6%`C. `99.9%`D. `22.2%` |
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Answer» Correct Answer - B `lambdaalpha1/V` |
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| 38. |
The kinetic energy of an electron is E when the incident wavelength is `lamda` To increase ti KE of the electron to 2E, the incident wavelength must beA. `2lambda`B. `lambda/2`C. `(hclambda)/(Elambda+hc)`D. `(2hclambda)/(Elambda+hc)` |
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Answer» Correct Answer - C `lambda_(1)/lambda=(omega+E)/(omega+2E)` |
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| 39. |
Ligth of wavelength `4000A^(@)` is incident on a metal surface of work function `2.5 eV`. Given `h=6.62xx10^(-34) Js,c=3xx10^(8) m//s`, the maximum `KE` of photoelectrons emitted and the corresponding stopping potential are respectivelyA. `0.6 eV,0.6 V`B. `2.5 eV,2.5 V`C. `3.1 eV,3.1 V`D. `0.6 eV,0.3 V` |
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Answer» Correct Answer - A `K.E=[(12400)/(lambda"in"A^(0))-omega_(0)]` |
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| 40. |
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. `(4lambda)/3`B. `4lambda`C. `6lambda1`D. `(8lambda)/3` |
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Answer» Correct Answer - B `(hC)/lambda=phi+eV`..(i) `(hC)/(2lambda)=phi+(eV)/3`..(ii) `3xxII-IrArr (3/2-1)(hc)/lambda=2phi rArr phi=(hc)/(4lambda)` `therefore lambda_(th)=4lambda` |
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| 41. |
A milliammeter in the circuit of a photocell measuresA. number of electrons released per secondB. energy of photonC. velocity of photoelectronsD. momentum of the photo electrons |
| Answer» Correct Answer - A | |
| 42. |
In a photocell bi chromatic light of wave length `2480 A^(0) ` and `6000 Aoverset(0)` are incident on a cathode whose workfunction is `4.8eV`.If a uniform magnetic field of `3xx10^(-5) T` exists parallel to the plate, find the radius of the circular path described by the photoelectron. (mass of electron is `9xx10^(-31)kg`) |
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Answer» `E_(1)=12400/lambda_(1)=12400/2480=5eV, E_(2)=12400/lambda_(2)=12400/6000=2.06eV,` As `E_(2) lt W_(0)` and `E_(1) lt W_(0)`, photo electric emission is possible only with `lambda_(1)`. Maximum `K.E` of emitted photo electrons `K=E_(1) - W_(0)=0.2eV` Photo electrons experience magnetic force and move along a circular path of radius `r=(mnu)/(Bq)=sqrt(2mK)/(Bq) rArr=5cm`. |
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| 43. |
Which of the following statements about the photoelectric effect, are trueA. greater the frequency of the incident light, greater is the stopping potential.B. greater the energy of photons is, the smaller is the stopping potential.C. greater the intensity of light is, greater is the cut off frequency.D. greater thefrequency of incident light is, greater is max kinetic energy of ejected electrons. |
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Answer» Correct Answer - A::D `KE_(max)` and stopping potantial increase with increase in frequency and threshold frequency depends only on the metal. |
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| 44. |
The work function of a metal is `2.5eV`. A monochromatic ligth of wavelength `300 Å` falls on itA. The maximum kinetic energy of ejected electron is `1.64eV` approximatelyB. The minimum kinetic energy of ejected electron is zeroC. The stopping potential is `1.64V` approximatelyD. Electrons can not be ejected |
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Answer» Correct Answer - A::B::C `KE_(max)=(hc)/lambda-phi` =`{((4.14xx10^(15))xx(3xx10^(8)))/(3000xx10^(-10))-2.5}eV` `1.64eV,K.E._(min)=zero` Stopping potential =`1/e((hc)/lambda-phi)=1.64V` |
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| 45. |
A monochromatic source of light operation at 200 W emits `4xx10^(20)` photons per second. Find the wavelength of the light `"(in" 10^(-7)m`). |
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Answer» Power`=N/t hv` Energy of photon`=E=P/((N/t))=200/(4xx10^(20))=5xx10^(-19)` `lambda=((6.62xx10^(34))xx(3xx10^(8)))/(5xx10^(-19))m=3.972A^(0)` |
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| 46. |
Consider a `20W` bulb emitting light of wavelength `5000Å` and shinning on a metal surface kept at a distance `2m`. Assume that the metal surface has work function of `2eV` and that each atom on the metal surface can be treated as a circular disk of radius `1.5Å`. (i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses] (ii) Will there be photoelectric emission? (iii) How much time would be required by the atomic disk to receive energy equal to work function `2eV`? (iv) How many photons would atomic disk receive within time duration calculated in (iii) above? (v) Can you explain how photoelectric effect was observed instantaneously? [Hint : Time calculated in part (iii) is from classical consideration and you may further take the target of surface area say `1cm^(2)` and estimate what would happen?]A. The number of photons emitted by the bulb per second. [Assume no other losses] is `5xx10^(19)s^(-1)`B. There will be photoelectric emissionC. Time required by the atomic disk to receive energy equal to work function `(2eV)` is `11.4s`D. The number of photons would atomic disk receive within time duration calculated in `(iii)` above is `2` |
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Answer» Correct Answer - A::B::C Given `P=20W,lambda=5000overset(0)(A)=5xx10^(-7) m`. `d=2m,phi_(0)=2eV,r=1.5oversetA=1.5xx10^(-10)m`. (A)Number of photons emitted by bulb per second is `n=P/((hc//lambda))=(Plambda)/(hc)=(20xx5xx10^(-7))/(6.62xx10^(-34)xx3xx10^(8))` (B)Energy of incident photon, `E=(hc)/lambda` =`((6.62xx10^(-34))(3xx10^(8)))/(5xx10^(-7)xx1.6xx10^(-19))=2.48eV`. As `E gt phi_(0)(248eV gt 2 eV)` hence photoelectric emission will take place. (C)Energy emitted by the bulb in time `Delta t=P Delta t` Energy falling on the disk i.e., `E=((PDeltat)/(4pid^(2)))(pir^(2))=((Pr^(2))/(4d^(2)))Delta t` According to given condition,`E=phi_(0)` `therefore ((Pr^(2))/(4d^(2)))Delta t=phi_(0),Deltat=(4phi_(0)d^(2))/(Pr^(2))` `Deltat=((4)(2xx1.6xx10^(-19))(2)^(2))/((20)(1.5xx10^(-10))^2)=11.4s` (D)Number of photons received by atomic disk in time `Deltat` is `N=n((pir^(2))/(4pid^(2)))xxDeltat=(nr^(2)Deltat)/(4d^(2))` `=(5xx10^(19)xx(1.5xx10^(-10))xx11.4)/(4(2)^(2))=0.8~~1` |
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| 47. |
`a)` Name the experiment for which the adjacent graph, showing the variation of intensity of scattered electrons with the angle of scattering `(theta)` was obtained. `b)` Also name the important hypothesis that was confirmed by this experiment. A. `A)` Davission and Germer experiment `B)` de Broglie hypothesisB. `A)` Photo electric effect `B)` de Broglie hypothesisC. `A)` Thermionic emission `B)` de Broglie hypothesisD. `A)` Photocell |
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Answer» Correct Answer - B A) Davission and Germer experiment B)de Broglie hypothesis |
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| 48. |
The photo electric proves that light consists ofA. PhotonsB. ElectronsC. Electromagnetic wavesD. Mechanical waves |
| Answer» Correct Answer - A | |
| 49. |
Insentity of light incident on a photo sensitive surface is doubled. ThenA. the number of emitted electrons is tripuledB. the number of emitted electrons is doubledC. the `K.E.` of emitted electrons is doubledD. the momentum of emitted electrons is doubled |
| Answer» Correct Answer - B | |
| 50. |
When orange light falls on a photo sensitive surface the photocurrent begins to flow. The velocity of emitted electrons will be more whwn surface is hit byA. red lightB. violet lightC. thermal radiationsD. radio waves |
| Answer» Correct Answer - B | |