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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.
| 301. |
The wavelength of the matter wave is independent ofA. MassB. VelocityC. MomentumD. Charge |
| Answer» Correct Answer - 4 | |
| 302. |
The graph between intensity of light falling on a metallic plate (l) with the current (i) generated isA. B. C. D. |
| Answer» Correct Answer - 2 | |
| 303. |
A particle A with a mass `m_(A)` is moving with a velocity v and hits a particle B (mass `m_(B)`) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat the collision as elastic. |
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Answer» As collision is elastic, hence laws of conservation of momentum and kinetic energy are obeyed. According to law of conservation of momentum, `" "1/2m_(A)v^(2)=1/2m_(A)v_(1)^(2)+1/2m_(B)v_(2)^(2)` `implies" "m_(A)(v-v_(1))-m_(B)v_(2)` According to law of conservation of kinetic energy, `" "1/2m_(A)v^(2)=1/2m_(A)v_(1)^(2)+1/2m_(B)v_(2)^(2)` `implies" "m_(A)(v-v_(1)^(2))=m_(B)v_(2)^(2)` `implies" "m_(A)(v-v_(1))(v+v_(1))=m_(B)v_(2)` Dividing Eq. (ii) by Eq. (i) we get, `" "v+v_(1)=v_(2)orv=v_(2)-v_(1)` Solving Eqs (i) and (iii), we get `" "v_(1)=((m_(A)-m_(B))/(m_(A)+m_(B)))vand v_(2)=((2m_(A))/(m_(A)+m_(B)))v` `" "lamda_("initial")=h/(m_(A)v)` `" "lamda_("final")=h/(m_(A)v_(1))=(h(m_(A)+m_(B)))/(m_(A)(m_(A)-m_(B))v)` `" "Deltalamda=lamda_("final")-lamda_("initial")=(h)/(m_(A)v)[(m_(A)+m_(B))/(m_(A)-m_(B))-1]` |
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| 304. |
A particle A with a mass `m_(A)` is moving with a velocity v and hits a particle B (mass `m_(B)`) at rest (one dimensional motion). Find the change the de-Broglie wavelength of the particle A. Treat the collision as elastic. |
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Answer» As collision is elastic, hence laws of conservation of momentum and kinetic energy are obeyed. According to law of conservation of momentum `m_(A)v+m_(B)0=m_(A)v_(1)+m_(B)v_(2) or m_(A)(v-v_(1))=m_(B)v_(2).......(i)` According to law of conservation of kinetic energy `1/2m_(A)v^(2)=1/2m_(A)v_(1)^(2)+1/2m_(B)v_(2)^(2) or m_(A)(v^(2)-v_(1)^(2))=m_(B)v_(2)^(2)` or `m_(A)(v-v_(1))(v-v_(1))=m_(B)v_(2)^(2).........(ii)` Dividing (ii) by (i), we get `v+v_(1)=v_(2) or v=v_(2)-v_(1)...(iii)` Solving (i) and (iii), we get `v_(1)=((m_(A)-m_(B))/(m_(A)+m_(B)))v` and `v_(2)=((2m_(A))/(m_(A)+m_(B)))v` `lambda_("initial")=h/(m_(A)v), lambda_("final")=h/(m_(A)v_(1))=(h(m_(A)+m_(B)))/(m_(A)(m_(A)-m_(B))v)` `:. Deltalambda=lambda_("final")-lambda_("initial")=h/(m_(A)v)[((m_(A)-m_(B)))/((m_(A)-m_(B)))-1]` |
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| 305. |
When a beam of accelerated electrons hits a target , a continuous `X` - ray spectrum is emitted from the target. Which of the following wavelength is absent in `X` - ray spectrum , if the `X` - ray tube is operating at `40,000 "volts"`?A. `0.25 Å`B. `0.5 Å`C. `1.5 Å`D. `1.0 Å` |
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Answer» Correct Answer - A `lambda_(min) = (12375)/(40,000) = 0.3 Å` , Hence wavelength less than `0.30 Å` is not possible . |
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| 306. |
A particle A with a mass `m_(A)` is moving with a velocity v and hits a particle B (mass `m_(B)`) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat the collision as elastic.A. `(h)/(2m_(A^(v)))[((m_(A)+m_(B)))/((m_(A)-m_(B)))-1]`B. `(h)/(m_(A^(v)))[((m_(A)-m_(B)))/((m_(A)+m_(B)))-1]`C. `(h)/(m_(A^(v)))[((m_(A)+m_(B)))/((m_(A)-m_(B)))-1]`D. `(2h)/(m_(A^(v)))[((m_(A)+m_(B)))/((m_(A)-m_(B)))+1]` |
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Answer» Correct Answer - C (c ) : According to law of conservation of momentum `m_(A)^(v)+m_(B)xx0m_(A)v_(A)+m_(B)v_(B)` or `m_(A)(v-v_(A))=m_(B)v_(B)` . . .(i) According to law of conservation of kinetic energy `(1)/(2)m_(A)v^(2)=(1)/(2)m_(A)^(2)+(1)/(2)m_(B)v_(B)^(2)` or `m_(A)(v^(2)-v_(A)^(2))=m_(B)v_(B)^(2)` or `m_(A)(v-v_(A))(v+v_(A))=m_(B)v_(B)^(2)` or `m_(A)(v-v_(A))(v+v_(A))=m_(B)v_(B)^(2)` . . . (ii) Dividing (ii) by (i), we get `v+v_(A)=v_(B) or v=v_(B)-v_(A)` . . . (iii) Solving (i) and (iii), we get `v_(A)=((m_(A)-m_(B))/(m_(A)+m_(B)))v and v_(B)=((2m_(A))/(m_(A)+m_(B)))v` `lamda_("initial")=(h)/(m_(A)v) and lamda_("final")=(h)/(m_(A)v_(A))=(h(m_(A)+m_(B)))/(m_(A)(m_(A)-m_(B))v)` `:.Deltalamda=lamda_("final")-lamda_("initial")=(h)/(m_(A)v)[((m_(A)+m_(B)))/((m_(A)+m_(B)))-1]` |
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| 307. |
The de Broglie wavelength associated with a ball of mass 150 g travelling at 30 m `s^(-1)` isA. `1.47xx1^(-34)m`B. `1.47xx1^(-16)m`C. `1.47xx1^(-19)m`D. `1.47xx1^(-31)m` |
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Answer» Correct Answer - A (a) : Mass of the ball, m=150 g = 0.15kg, Speed of the ball, v = 30 m `s^(-1)` Momentum, P = mv = `0.15xx30=4.5kg" m s"^(-1)` de Broglie wavelength, `lamda=(h)/(p)=(6.63xx10^(-34))/(4.5)=1.47xx10^(-34)m` |
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| 308. |
What is the de-Broglie wavelength associated with (a) an electron moving with speed of `5.4xx10^6ms^-1`, and (b) a ball of mass 150g travelling at `30.0ms^-1`? `h=6.63xx10^(-34)Js`, mass of electron `=9.11xx10^(-31)kg`. |
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Answer» For the electron , `m=9.11xx10^(-31) , v=5.4xx10^6ms^-1` `lambda=h/(mv)=(6.63xx10^(-34))/((9.11xx10^(-31))xx(5.4xx10^6))=0.135xx10^-9m =0.135nm` (b) For a ball , `m=150g=150xx10^-3kg, v=30.0ms^-1` `lambda=h/(mv)=(6.63xx10^(-34))/((150xx10^-3)xx30.0)=1.47xx10^(-34)m` |
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| 309. |
The electric field associated with a monochromataic beam of light becomes zero, `2.4xx10^(15)` times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0eV, `h=6.63xx10^(-34)Js`. |
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Answer» Correct Answer - `2.97 eV` In one complete vibration twice the electric field becomes zero, so frequency of the incident light, `v=1/2xx2.4xx10^(15)=1.2xx10^(15)cps`. Max K.E. `=hv-phi_(0)` `=(6.63xx10^(-34)xx1.2xx10^(15))/(1.6xx10^(-19))-2.0` `=2.97 eV` |
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| 310. |
Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.65 eV to be ejected from a metal surface whose work function is 2.65 eV. What will be the maximum kinetic energy (in eV) with which this same radiation ejects electrons from another metal whose work function is 2.15eV. |
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Answer» Correct Answer - `1.05 eV` Here, `K_(max 1)=0.65 eV , phi_(0)=0.65+2.65=3.20 eV` Case (ii) `(K_(max))_(2)=hv-phi_(02) =3.20-2.15=1.05 eV` |
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| 311. |
Light of frequency `7.21xx10^(14)` Hz is incident on metal surface. Electrons with a maximum speed of `6.0xx10^(5)m//s` are ejected from the surface. What is the threshold frequency for photoemission of electrons ? |
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Answer» Given, `v =7.21xx10^(14) Hz, m = 9.1xx10^(-31)kg, upsilon_("max")=6xx10^(5)m//s` `KE_("max")=(1)/(2)m upsilon_("max")^(2)= hv-hv_(0)= h(v-v_(0))` `rArr v -v_(0)=((1)/(2)mv_("max")^(2))/(h)=(1)/(2)xx((9.1xx10^(-31))xx(6xx10^(5))xx(6xx10^(5)))/(6.63xx10^(-34)) rArr v -v_(0)=2.47xx10^(14)` `rArr v_(0)= v-2.47xx10^(14)=7.21xx10^(14)-2.47xx10^(14) therefore v_(0) = 4.74xx10^(14)Hz`. |
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| 312. |
Light of frequency `7.21xx10^(14)Hz` is incident on a metal surface. Electrons with a maximum speed of `6.0xx10^(5)ms^(-1)` are ejected from the surface. What is the threshold frequency for photoemission of electrons? `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`. |
| Answer» `1/2mv_(max)^(2)=hv-hv_(0) or v_(0) =v-(mv_(max)^(2))/(2h)=7.21xx10^(14)-((9.1xx10^(-31))xx(6xx10^(5))^(2))/(2xx(6.63xx10^(-34)))=4.74xx10^(14) Hz` | |
| 313. |
The maximum wavelength for which an em wave can ejected electrons form a platinum surface is 195 nm. When radiations with a wavelength of 131nm shines on the surface. What is the maximum speed of the ejected electrons? [Use `h=6.63xx10^(-34)Js`] `m_(e)=9.1xx10^(-31)kg`. |
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Answer» Correct Answer - `.105xx10^(6)ms^(-1)` Here, `lambda_(0)=195nm=195xx10^(-9)m`, `lambda=131xx10^(-9)m` `1/2mv_(max)^(2)=(hc)/lambda-(hc)/(lambda_(0))` `=(6.63xx10^(-34)xx3xx10^(8))/(131xx10^(-9))-(6.63xx10^(-34)xx3xx10^(8))/(195xx10^(-9))` `=0.152xx10^(-17)-0.102xx10^(-17)J` `v_(max)=[2/mxx0.050xx10^(-17)]^(1//2)` `=[2/(9.1xx10^(-31))xx0.050xx10^(-17)]^(1//2)` `=1.05xx10^(6)ms^(-1)` |
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| 314. |
The potential difference applied to an X-ray tube is 5k V and the current through it is 3.2 mA. Then, the number of electrons striking the target per second is. (a) `2xx10^16` (b) `5xx10^6` (c ) `1xx10^17` (d) `4xx10^15`.A. `2 xx 10^(16)`B. `5 xx 10^(16)`C. `1 xx 10^(17)`D. `4 xx 10^(15)` |
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Answer» Correct Answer - A `i = (Ne)/(t) rArr (N)/(t) = (i)/( e) = (3.2 xx 10^(-3))/(1.6 xx 10^(-19)) = 2 xx 10^(16)//sec` |
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| 315. |
Assertion : The specific charge of positive rays is not constant. Reason : The mass of ions varies with speed.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - B The specific charge `( e//m)` of the positive rays is not universal constant because these rays may consists of ions of different element. |
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| 316. |
Assertion : The specific charge for positive rays is a characteristic constant. Reason : The specific charge depends on charge and mass of positive ions present in positive rays .A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - B Specific charge of a positive ion corresponding to one gas is fixed but it is different for different gases. |
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| 317. |
Assertion : Isotope is possible because of the using a mass spectrometer. Reason : Separation of isotope is possible because of the difference in electron number of isotope.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not correct explanation of the assertion.C. If assertion is true but the reason is false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - C The atomic number (number of electrons or protons ) remains same in isotope. Isotope of an element can be separated on account of their different atomic weight by using mass spectrograph. |
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| 318. |
Assertion : The specific charge of positive rays is not constant. Reason : The mass of ions varies with speed.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not correct explanation of the assertion.C. If assertion is true but the reason is false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - B The specific charge `(e//m)` of the positive rays is not universal constant because these rays may consist of ions of different elements. |
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| 319. |
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. What is the energy per photon associated with the sodium light ? |
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Answer» Given, `P = 100 W, lambda = 589 nm = 589xx10^(-9)m, h = 6.63xx10^(-34)J-s, c = 3xx10^(8)m//s` `E =(hc)/(lambda)=(6.63xx10^(-34)xx3xx10^(8))/(589xx10^(-9))=3.38xx10^(-19)J = (3.38xx10^(-19))/(1.6xx10^(-19))eV = 2.11 eV`. |
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| 320. |
A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589nm. (a) What is energy associated per photon with the sodium light? (b) At what rate are photons delivered to the sphere?A. `3xx10^(15)`B. `3xx10^(10)`C. `3xx10^(20)`D. `3xx10^(19)` |
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Answer» Correct Answer - C (c ) : Here `lamda=589xx10^(-9)m,h=6.63xx10^(-34)J` s P = 100W Energy of a photon, `E=(hc)/(lamda)=(6.63xx10^(-34)xx3xx10^(-19))/(589xx10^(-9))=3.38xx10^(-19)J` Number of photons delivered per second `n=(P)/(E)=(100)/(3.38xx10^(-9))=3xx10^(20)` |
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| 321. |
A 100 W sodium lamp radiates radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. At what rate are the photons delivered to the sphere ? |
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Answer» Given, `P = 100 W, lambda = 589 nm = 589xx10^(-9)m, h = 6.63xx10^(-34)J-s, c = 3xx10^(8)m//s` No. of photons deliverd per second, `N = (P)/(E )=(100)/(3.38xx10^(-19))=3xx10^(20)` photons/s. |
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| 322. |
A beam of light has three wavelengths `4144 Å`, `4972 Å` and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects on electron. Calculate the number of photo electrons liberated in two seconds. |
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Answer» Here, `phi_(0) =2.3eV=2.3xx1.6xx10^(-19)J` Thereshold wavelength, `lambda_(0)=(hc)/(phi_0)=((6.63xx10^(-34))xx(3xx10^(8)))/(2.3xx1.6xx10^(-19))` `=5.404xx10^(-7)=5404Å` This shows that the beam of light of wavelength `4144Å and 4972Å` will emit photoelectrons from the metal surface. The enery incident on the surface for each wavelength of light =intensity of each wavelengthx area of surface `=((3.6xx10^(-3)))/3xx(1.0xx10^(-4))=1.2xx10^(-7)wat t` Energy incident on the surface for each wavelength of light in 2 second is `E=(1.2xx10^(-7))xx2=2.4xx10^(-7)joule` No. of photons of wavelength `lambda` are `n=E/(hc//lambda)=(Elambda)/(hc)` No. of photons `n_(1)` due to wavelength `4144Å` is `n_(1)=((2.4xx10^(-7))xx(4144xx10^(-10)))/((6.63xx10^(- 34))xx(3xx10^(8)))=0.5xx10^(12)` No. of photons `n_(2)` due to wavelength `4972Å` is `n_(2)=((2.4xx10^(-7))xx(4972xx10^(-10)))/((6.63xx10^(-34))xx(3xx10^(8)))=0.576xx10^(12)` Total number of photoelectrons emitted =total number of photons falling which cause photon emission `=n_(1)+n_(2)=0.5xx10^(12)+0.576xx10(12)` `=1.075xx10^(12)` |
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| 323. |
The work function for tungsten and sodium are `4.5 eV` and `2.3 eV` respectively . If the threshold wavelength `lambda` for sodium is `5460 Å`, the value of `lambda` for tungsten isA. `5893 Å`B. `10683 Å`C. `2791 Å`D. `528 Å` |
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Answer» Correct Answer - C Since `W_(0) = (hc )/(lambda_(0)), :. ((W_(0))_ T)/((W_(0))_(Na)) = (lambda_(Na))/(lambda_(T))` or `lambda_(T) = ((lambda_(Na)) xx (W_(0))_(Na))/((W_(0))T) = ( 5460 xx 2.3)/(4.5) = 2791 Å` |
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| 324. |
The specific charge of a proton is `9.6xx10^(7)"C kg"^(-1)`. The specific charge of an alpha particle will beA. `9.6xx10^(7)"C kg"^(-1)`B. `19.2xx10^(7)"C kg"^(-1)`C. `4.8xx10^(7)"C kg"^(-1)`D. `2.4xx10^(7)"C kg"^(-1)` |
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Answer» Correct Answer - C (c ) : For proton, Specific charge `=(e)/(m)=9.6xx10^(7)"C kg"^(-1)` For alpha particle, Specific charge `=(2e)/(4m)=(1)/(2)(e)/(m)=(1)/(2)xx9.6xx10^(7)` `=4.8xx10^(7)"C kg"^(-1)` |
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| 325. |
Which of the following statements is correct regarding the photoelectric experiment ?A. The photocurrent increases with intensity of light.B. Stopping potential increases with increase in intensity of incident light.C. The photocurrent increases with increase in frequency.D. All of these |
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Answer» Correct Answer - A (a) : The photocurrent increases linearly with intensity of the light, but is independent of its frequency. The stopping potential increases linearly with the frequency of the incident light, but is independent of its intensity. |
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| 326. |
Thermionic emissions are related toA. conductionB. convectionC. radiationD. none of these |
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Answer» Correct Answer - B (b) : Thermionic emissions are emission of electrons, they transport current. |
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| 327. |
In photoelectic effect, the photocurrentA. depends both on intensity and frequency of the incident light.B. does not depand on frequency of incident light but depends on intensity of the incident light.C. decreases with increase in frequency of incident light.D. increases with increase in frequency of incident light. |
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Answer» Correct Answer - B (b) : Photoelectric current depends on the intensity of the incident light and is independent of the frequency of incident light. |
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| 328. |
The phenomenon of photoelectric emission was discovered in 1887 byA. Albert EinsteinB. Heinrich HertzC. Wilhelm HallwachsD. Philipp Lenard |
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Answer» Correct Answer - B (b) : The phenomenon of photoelectric emission was discovered by Heinrich Hertz in 1887. |
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| 329. |
In an experiment on photoelectic effect, the graph as shown in fig. were obtained between the photoelectric current (I) and the anode potential (V). Name the characterstic of the incident radiation that was kept constant in this experiment. |
| Answer» As the value of stopping potential is same for all the curves, so the frequency of incident radiations is kept constant but their intensity is different. | |
| 330. |
Assertion : The phenomenon of `X` - ray production is basically inverse of photoelectric effect. Reason : `X` - rays are electromagnetic waves.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - B In photoelectric effect , the photon falling on some matter and its energy is transferred to an electron of the matter. In `X` - ray production , photons are produced which get energy from energetic electrons ionizing the inner shells of the target which in turn cause a cascade of emission lines. |
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| 331. |
Dual nature of radiation is shown byA. Diffraction and reflectionB. Refraction and diffractionC. Photoelectric effect aloneD. Photoelectric effect and diffraction |
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Answer» Correct Answer - D `{["Phot oe l ectric ef f ect rarr Particl e nature"],["Di f f r a c t ion" rarr "Wave nature"]}` Dual nature |
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| 332. |
The de - Broglie wavelength `lambda`A. is proportional to massB. is inversely to massC. Inversely proportional to linear momentumD. does not depend on linear momentum |
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Answer» Correct Answer - C The de - Broglie wavelength is `lambda = (h )/(|P|) = (h)/(|I|) rArr lambda prop (1)/(|I|)` |
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| 333. |
Find the typical de-Broglie wavelength associated with a H-atom in helium gas at room temperature `(27^(@)C)` and 1atm pressure, and compare it with the mean separation between two atom under these conditions. |
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Answer» Given `T = 27 + 273 = 300 K, K = 1.38 xx10^(-23) `J/mol/K P =1 atm `= 1.01 xx 10^(5)` Pa Mass of atom` =("Atomic weight")/("Avagadro number") = 4/(6xx 10^(23)) g = 4/(6xx10^(26))kg` ` lambda - h/(sqrt(3"mKT"))= (6.63 xx10^(-34))/(sqrt(3 xx 4/(6xx10^26)xx 1.38 xx10^(-23) xx300) )= 0.73 xx10^(-10)` m Now , PV = Rt = KNT, `V/N = (KT)/P` Mean seperation r = `[(V)/N]^(1/3)= [(KT)/P]^(1//3) = [ (1.38xx10^(-23)xx300)/(1.01 xx10^(5))]^(1//3)` ` r = 3.4 xx 10^(-9) m , lambda /r = (0.73xx10^(-10))/(3.4 xx10^(-9))= 0.021 ` We can see that the wavelength `lambda` with mean seperation r, it can be observed `(r gt gt lambda)` that seperation is larger than wavelength . |
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| 334. |
An electron and a photon have the same de Broglie wavelength. Which one of these has higher kinetic energy?A. ZeroB. InfinityC. Equal to the kinetic energy of the protonD. Greater than the kinetic energy of the proton |
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Answer» Correct Answer - D `lambda = (h)/(sqrt(2 m E)) rArr E prop (1)/(sqrt(m))` `(lambda = constant)` `because m_(e ) lt m_(p)` So `E_(e) gt E_(p)` |
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| 335. |
The de - Broglie wavelength of a particle moving with a velocity `2.25 xx 10^(8) m//s` is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is (velocity of light is `3 xx 10^(8) m//s`A. `1//8`B. `3//8`C. `5//8`D. `7//8` |
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Answer» Correct Answer - B `K_("particle") = (1)/(2) mv^(2)` also `lambda = (h)/(mv)` `rArr K_("particle") = (1)/(2) ((h)/(lambda v)).v^(2) = (vh)/(2 lambda)` ….(i) `K_(photon) = (hc)/(lambda)` ….(ii) `:. (K_("particle"))/(K_("particle")) = (v)/( 2c) = ( 2.25 xx 10^(8))/( 2 xx 3 xx 10^(8)) = (3)/(8)` |
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| 336. |
An electron and a photon have the same de Broglie wavelength. Which one of these has higher kinetic energy? |
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Answer» For electron. Let `lambda` be the de -Broglie wavelength of electron. K.E. of electron, `E_1=1/2mv^2 or mv^2=2E_1` or `mv=sqrt(2E_1m)` As `lambda=h/(mv) :. lambda=h/(sqrt(2E_1m))` or `E_1=(h^2)/(2 lambda^2m)` For photon of wavelength `lambda`, Energy is given by, `E_2=(hc)/lambda` `:. (E_2)/(E_1)=(hc)/lambdaxx(2lambda^2m)/(h^2)=(2clambdam)/h` `=(2xx3xx10^8xx10^(-10)xx8xx10^(-31))/(6.6xx10^(-34))` `=(2xx3xx9xx10)/6.6=90/1.1gt1` ltbegt Hence, `E_2gtE_1` It means kinetic energy of photon is greater than that of electron. |
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| 337. |
A proton and an electron have same d-Broglie wavelength which of them moves fast and which possess more K.E. Justify your answer. |
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Answer» Kinetic energy K of a particle of mass m having Momentum p is , `K=1/2 (p^2)/m or p=sqrt(2mK)` De-Broglie wavelength, `lambda=h/p=h/(sqrt(2mK))` `:. P=h/lambda` ....(i) and `K=(h^2)/(2mlambda^2)`.....(ii) If `lambda` is constant, then from (i), p=a constant i.e., `m_pv_p=m_ev_e` or `(v_p)/(v_e)=(m_e)/(m_p) lt 1` or `v_p lt v_e` If `lambda` is constant, the form (ii), `Kprop1//m` `(K_p)/(K_e)=(m_e)/(m_p) lt 1` or `K_p lt K_e` It means the velocity of electron is greater than that of proton. Kinetic energy of electron is greater than that of proton. |
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| 338. |
How fast does a proton have to be moving with order to have the same de-Broglie wavelength as an electron that is moving with a speed of `5.5xx10^(6)ms^(-1)`? Mass of proton`=1.67xx10^(-27)kg`. |
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Answer» Let `v_(p)` be the speed of proton for which de-Broglie wavelength of proton =de-Broglie wavelength of moving electron `h/(m_(p)v_(p))=h/(m_(e)v_(e))` of `v_(p)=v_(e)xx(m_(e))/(m_(p))` `=(5.5xx10^(6))xx((9.1xx10^(-31))/(1.67xx10^(-27)))` `2.45xx10^(3)ms^(-1)` |
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| 339. |
An electron in hydrogen like atom is in excited state 2t has a total energy of -3.4eV. Calculate (a) the kinetic energy (b) the de-Broglie wavelength of the electron. [Given `h=6.63xx10^(-34)Js`] |
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Answer» (a) Total energy of electron in a hydrogen like atom is given by `E=KE+PE` Also, `KE=1/2(PE) or PE=-2KE` `:. E=KE-2KE=-KE=(-3.4eV)` `=3.4eV` (b) De-Broglie wavelength of electron is `h=h/(mv)=h/([2m(KE)]^(1//2))` `=(6.63xx10^(-34))/([2xx(9.1xx10^(-31))xx(3.4xx1.6xx10^(-19))]^(1//2))` `=(6.63xx10^(-34))/(9.95xx10^(-25))=6.66xx10^(-10)m=6.66Å` |
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| 340. |
Fraunhofer line of the solar system is an example ofA. line absorption spectrumB. band absorption spectrumC. line emission spectrumD. band emission spectrum |
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Answer» Correct Answer - A The atoms in the chromosphere absorb certain wavelengths of light coming from the photosphere. This gives rise to absorption lines. |
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| 341. |
A caesium photocell , with a steady potential difference of `60 V` across , is alluminated by a bright point source of light `50 cm` away. When the same light is placed `1 m` away the photoelectrons emitted from the cellA. Are one quarter as numerousB. Are half as numerousC. Each carry one quarter of their previous momentumD. Each carry one quarter of their previous energy |
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Answer» Correct Answer - A Number of photo electrons `(N) prop Intensity prop (1)/(d^(2)) rArr (N_(1))/(N_(2)) = ((d_(2))/(d_(1)))^(2)` `rArr (N_(1))/(N_(2)) = ((100)/(50))^(2) = (4)/(1) rArr N_(2) = (N_(1))/(4)`. |
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| 342. |
Ultraviolet light of wavelength 300nn and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearlyA. `9.61 xx 10^(14) per sec`B. `4.12 xx 10^(13) per sec`C. `1.51 xx 10^(12) per sec`D. `2.13 xx 10^(11) per sec` |
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Answer» Correct Answer - C Intensity of light `I = ( "Watt")/("Area") = ( nhc)/( A lambda) rArr "Number of photon" , n = (I A lambda)/(hc)` `:.` Number of photo electron ` = (1)/(100) xx (IA lambda)/(hc)` `= (1)/(100) ( 1 xx 10^(-4) xx 300 xx 10^(-9))/(6.6 xx 10^(-34) xx 3 xx 10^(8)) = 1.5 xx 10^(12)` |
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| 343. |
The approximate wavelength of a photon of energy `2.48 eV` isA. `500 Å`B. `5000 Å`C. `2000 Å`D. `1000 Å` |
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Answer» Correct Answer - B By using `E( eV) = (12375)/(lambda (Å))` `rArr lambda = (12375)/(2.48) = 4989.9 Å ~~ 5000 Å` |
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| 344. |
The momentum of a photon in an X - ray beam of `10^(-10) metre` wavelength isA. `1.5 xx 10^(-23) kg - m//sec`B. `6.6 xx 10^(-24) kg - m//sec`C. `6.6 xx 10^(-44) kg - m//sec`D. `2.2 xx 10^(-52) kg - m//sec` |
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Answer» Correct Answer - B Momentum of photon `p = (h)/(lambda) = (6.6 xx 10^(-34))/(10^(-10)) = 6.6 xx 10^(-24) kg - m//sec` |
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| 345. |
Ultraviolet light of wavelength `2271Å` from a 100W mercury souce irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3V, estimate the work function of the metal. How would the photocell respond to a high intensity `(~10^(5)Wm^(-2))` red light of wave length `6328Å` produced by a He-Ne laser? |
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Answer» Here, `lambda=2271Å=2271xx10^(-10)m, V_(0)=1.3V, phi_(0)=?` we know, `eV_(0)=hv-phi_(0) or phi_(0)=hv-eV_(0)=(hc)/lambda-eV_(0) = (6.63xx10^(-34)xx3xx10^(8))/(2271xx10^(-10))-1.6xx10^(-19)xx1.3` `phi_(0)=6.7xx10^(-19)J= (6.7xx10^(-19))/(1.6xx10^(-19))eV=4.2 eV` If `V_(0)` is the frequency, then `hv_(0) =(phi_(0))/h=(6.7xx10^(-19))/(6.63xx10^(-34)) =1.0xx10^(15)Hz` For red light, `lambda_(r)=6328Å=6238xx10^(-10)m`. we have, frequency `v_(r)=c/(lambda_(r))=(3xx10^(8))/(6238xx10^(-10))=4.74xx10^(14)Hz`. Since, `v_(r) lt v_(0)`, therefore, the photocell will not respond to this red light, however strong its intensity may be. |
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| 346. |
Monochromatic radiation of wave length 640.2 nm `(1nm=10^(-9)m)` from a neon lamp irradiates a photosenstive material made of calcium or tungsten. The stopping voltage is measured to be 0.54V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photocell. Predict the new stopping voltage. |
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Answer» Here, For neon lamp, `lambda=640.2nm =640.2xx10^(-9)m, V_(0)=0.54V. We know, eV_(0)=(hc)/lambda-phi_(0)` `:.` Work function, `phi_(0)=(hc)/lambda-eV_(0)=(6.63xx10^(-34)xx3xx10^(8))/(640.2xx10^(-9))-1.6xx10^(-19)xx0.54` `=3.1xx10^(-19)-0.864xx10^(-19)J=2.236xx10^(-19)J= (2.236xx10^(-19))/(1.6xx10^(-19))eV=1.4eV` For iron lamp, `lambda=427.2nm =427.2xx10^(-9)m`, `:. eV_(0)=(hc)/lambda-phi_(0)` `=(6.63xx10^(-34)xx3xx10^(8))/(427.2xx10^(-9))-2.236xx10^(-19)` `=4.649xx10^(-19)-2.236xx10^(-19)J=2.413xx10^(-19)J` `:.` Stopping potential, `V_(0)=(2.413xx10^(-19))/e= (2.413xx10^(-19))/(1.6xx10^(-19))eV=1.55V` |
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| 347. |
Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured material. The threshold frequency of the material isA. `2.5 xx 10^(15) Hz`B. `4 xx 10^(15) Hz`C. `5 xx 10^(15) Hz`D. `1.6 xx 10^(15) Hz` |
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Answer» Correct Answer - D Energy released from emission of electron `E = (-3.4) - (-13.6) = 10.2 eV` From photo electric equation. Work function `phi = E = eV = hv` `v = (E - eV)/(h) = ((10.2 - 3.57)e)/(6.67 xx 10^(-34))` `v = (6.63 xx 1.6 xx 10^(-19))/(6.67 xx 10^(-34)) = 1.6 xx 10^(15) Hz` |
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| 348. |
Photons with energy `5 eV` are incident on a cathode `C` in a photoelectric cell . The maximum energy of emitted photoelectrons is `2 eV`. When photons of energy `6 eV` are incident on `C` , no photoelectrons will reach the anode `A` , if the stopping potential of `A` relative to `C` isA. ` - 1 V`B. `- 3 V`C. `+ 3 V`D. `+ 4 V` |
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Answer» Correct Answer - B `eV_(s) = (1)/(2) mv_(max)^(2) = hv - phi_(0)` ` 2 = 5 - phi_(0) rArr phi_(0) = 3 eV` In second case `eV_(s) = 6 - 3 = 3 eV rArr V_(s) = 3 V` `:. V_(AC) = - 3 V` |
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| 349. |
The photoelectric cut-off voltage in certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ? |
| Answer» Give, `V_(0)=1.5 V, e = 1.6xx10^(-19)C, KE_("max")=eV_(0)=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J`. | |
| 350. |
The photoelectric cut off voltage in a certain experiment is 1.5V. What is the maximum kinetic energy of photoelectrons emitted? `e=1.6xx10^(-19)C`.A. 2.4 eVB. 1.5 eVC. 3.1 eVD. 4.5 eV |
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Answer» Correct Answer - B (b) : Here, `V_(0)=1.5V`, Maximum kinetic energy `=eV_(0)=1.5eV` |
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