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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.
| 201. |
Stopping potential for photo electronsA. Does not depend on the frequency of the incident lightB. Does not depend upon the nature of the cathode materialC. Depends on both the frequency of the incident light and nature of the cathode materialD. Depends upon the intensity of the incident light |
| Answer» Correct Answer - 3 | |
| 202. |
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut - off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. |
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Answer» Given, `lambda = 488 nm = 488xx10^(-9)m, V_(0)=0.38 V, e=1.6xx10^(-19)c, h=6.63xx10^(-34)J-s` `c=3xx10^(8)m//s rArr KE = eV_(0)=(hc)/(lambda) - phi_(0) rArr 1.6xx10^(-19)xx0.38=(6.63xx10^(-34)xx3xx10^(8))/(488xx10^(-9))- phi_(0)` `6.08xx10^(-20)=40.75xx10^(-20)-phi_(0) rArr phi_(0) = (40.75-6.08)xx10^(-20)=34.67xx0^(-20)J` `= (34.67xx10^(-20))/(1.6xx10^(-19))eV therefore phi_(0)=2.17 eV`. |
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| 203. |
The maximum wavelength of radiation that can produce photoelectric effect in a certain metal is 200 nm . The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will beA. 12.4 eVB. 6.2 eVC. 100 eVD. 200 eV |
| Answer» Correct Answer - 2 | |
| 204. |
Which of the following pheniomeana exhibits particle nature of light ?A. Photoelectric effectB. InterferenceC. RefractionD. Polarization |
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Answer» Correct Answer - A (a) : Particle nature of light was established by photoelecric effect. |
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| 205. |
If the momentum of an electron is changed by `Delta p` , then the de - Broglie wavelength associated with it changes by `0.50 %`. The initial momentum of the electron will beA. `(Delta p)/(200)`B. `(Delta p)/(199)`C. `199 Delta p`D. `400 Delta p` |
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Answer» Correct Answer - C `lambda = (h)/(p) rArr lambda - (0.5)/(100) lambda = (h)/( p + Delta p)` `rArr (199 lambda)/(200) = (h)/(p + Delta p) = (199)/(200) (h)/(p)` `rArr p + Delta p = (2000)/(199) p rArr p = 199 Delta p` |
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| 206. |
What is the de-Broglie wavelength of (a) a bullet of mass `0.040kg` traveling at the speed of `1.0kms^(-1)`. (b) a ball of mass `0.060 kg` moving at a speed of `1.0ms^(-1)` and (c) a dust particle of mass `1.0xx10^(-9) kg` drifting with a speed of `2.2 ms^(-1)`? `h=6.63xx10^(-34)Js`. |
| Answer» Given, for ball m = 0.060 kg and `upsilon = 1 m//s rArr lambda = (h)/(m upsilon)= (6.63xx10^(-34))/(0.060xx1)=1.1xx10^(-32)m` | |
| 207. |
Calculate the de-Broglie wavelength of a 0.20 kg ball moving with a speed of 15 m/s |
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Answer» `lambda=h/p=h/"mv"` `=((6.6xx10^(-34)J.s))/((0.20 kg)(15 m//s))` `=2.2xx10^(-34)` m |
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| 208. |
The wavelength of de - Broglie wave is `2 mu m` , then its momentum is `( h = 6.63 xx 10^(-34 J-s)`A. `3.315 xx 10^(-28) kg - m//s`B. `1.66 xx 10^(-28) kg - m//s`C. `4.97 xx 10^(-28) kg-m//s`D. `9.9 xx 10^(-28) kg - m//s` |
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Answer» Correct Answer - A `lambda = (h)/(p) rArr p = (h)/(lambda) = (6.63 xx 10^(-34))/(2 xx 10^(-6))` `= 3.31 xx 10^(-28) kg - m//sec` |
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| 209. |
A proton and and `alpha `- particle are accelerated through a potential difference of `100 V`. The ratio of the wavelength with the proton to that associated to that associated with an `alpha` - particle isA. `sqrt(2) : 1`B. `2 : 1`C. `2 sqrt(2) : 1`D. `(1)/( 2 sqrt(2)) : 1` |
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Answer» Correct Answer - C `lambda = (h)/(sqrt(2 m QV)) rArr lambda prop (1)/(sqrt(mQ)) rArr (lambda_(p))/(lambda_(alpha)) = sqrt((m_(alpha) Q_(alpha))/( m_(p) Q_(p)))` `rArr sqrt((4m_(p) xx 2 Q_(p))/( m_(p) xx Q_(p))) = 2 sqrt(2)` |
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| 210. |
When the mkomentum of a proton is changed by an amount `p_(0)`, the corresponding change in the de-Broglie wavelength is found to be `0.25%`. Then, the original momentum of the proton was |
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Answer» `lambda prop 1/p` `rArr (Deltap)/p= -(Deltalambda)/lambda` `rArr p_0/p=0.25/100` `rArr p=400p_0` |
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| 211. |
A proton and and `alpha `- particle are accelerated through a potential difference of `100 V`. The ratio of the wavelength with the proton to that associated to that associated with an `alpha` - particle is |
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Answer» `lambda=h/sqrt(2mQV)` `rArr lambda prop 1/sqrt(mQ)` `rArr lambda_p/lambda_alpha=sqrt((m_alpha Q_alpha)/(m_p Q_p))` `=sqrt((4m_pxx2Q_p)/(m_pxxQ_p))` =`2sqrt2` |
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| 212. |
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 isA. `1`B. `(1)/(4)`C. `(1)/(2)`D. `2` |
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Answer» Correct Answer - C When a charged particle (charge `q , mass m`) enters perpendicularly in a magnetic field `(B)` than , radius of the path described by it `r = (mv)/(q^(B)) rArr mv = qBr`. Also de - Broglie wavelength `lambda = (h)/(mv)` `rArr lambda = (h)/( qBr) rArr (lambda_(alpha))/(lambda_(p)) = (q_(p) r_(p))/(q_(alpha) r_(alpha)) = (1)/(2)` |
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| 213. |
The work function of a substance is `4.0 eV`. The longest wavelength of light that can cause photo electron emission from this substance is approximately. (a) `540 nm` (b ) `400nm` (c ) `310 nm` (d) `220 nm`A. `540 nm`B. `400 nm`C. `310 nm`D. `220 nm` |
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Answer» Correct Answer - C `lambda_(0) = (hc)/(W_(0)) = (12400)/(4) = 3100 Å = 310 nm` |
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| 214. |
The duration of a laser pulse is `10^(-8)` s. The uncertainly in its energy will beA. `6.6xx10^(-26)`JB. `6.6xx10^(-42)` JC. `6.6xx10^(-34)` JD. `1/6.6xx10^26` J |
| Answer» Correct Answer - 1 | |
| 215. |
If the velocity of an electron is doubled, its de-Broglie frequency will beA. Be halvedB. Remain sameC. Be doubledD. Become four times |
| Answer» Correct Answer - 3 | |
| 216. |
When the electromagnetic radiations of frequencies `4 xx 10^(15) Hz` and `6 xx 10^(15) Hz` fall on the same metal, in different experiments, the ratio of maximum kinetic energy of electrons liberated is `1 : 3`. The threshold frequency for the metal isA. `2xx10^15` HzB. `1xx10^15` HzC. `3xx10^15` HzD. `1.67xx10^15` Hz |
| Answer» Correct Answer - 3 | |
| 217. |
The momentum of a photon is `2 xx 10^(-16) gm - cm//sec`. Its energy isA. `0.61xx10^(-26)` ergB. `2.0xx10^(-26)` ergC. `6xx10^(-6) ` ergD. `6xx10^(-8)` erg |
| Answer» Correct Answer - 3 | |
| 218. |
The work function of a certain metal is 2.3 eV. If light of wave number `2 xx 10^(6)m^(-1)` falls on it, the kinetic energies of fastest and slowest ejected electron will be respectively.A. 2.48 eV , 0.18 eVB. 0.18 eV, zeroC. 2.30 eV, 0.18 eVD. 0.18 eV , 0.18 eV |
| Answer» Correct Answer - 2 | |
| 219. |
The momentum of a photon is `33 xx 10^(- 29) kg - m//sec`. Its frequency will beA. `3xx10^3` HzB. `6xx10^3` HzC. `7.5xx10^12`HzD. `1.5xx10^13` Hz |
| Answer» Correct Answer - 4 | |
| 220. |
A red bulb and violet bulb of equal power emits `n_(R )` and `n_(v)` number of photons in a given time, thenA. `n_R=n_V`B. `n_R gt n_V`C. `n_R lt n_V`D. `n_R ge n_V` |
| Answer» Correct Answer - 2 | |
| 221. |
Specific heat of water is `4.2 J//g.^(@)C`. If light of frequency `3 xx 10^(29)Hz` is used to heat 400 g of water from `20^(@)C` to `40^(@)C`, the number of moles of photons needed will beA. `1.69xx10^29`B. `1.69xx10^28`C. `2.80xx0^4`D. `2.80xx10^5` |
| Answer» Correct Answer - 3 | |
| 222. |
Two identical photocathodes receive light of frequencies `f_1` and `f_2`. If the velocities of the photo electrons (of mass m ) coming out are respectively `v_1` and `v_2` thenA. `f_1-f_2=(m(v_1^2-v_2^2))/(2h)`B. `f_1+f_2=(m(v_1^2-v_2^2))/(2h)`C. `f_1-f_2=(m(v_1^2+v_2^2))/(2h)`D. `f_1+f_2=(m(v_1^2+v_2^2))/(2h)` |
| Answer» Correct Answer - 1 | |
| 223. |
Kinetic energy with which the electrons are emitted from the metal surface due to photoelectric effect isA. Independent of the intensity of illuminationB. Independent of the frequency of lightC. Inversely proportional to the intensity of illuminationD. Directly proportional to the intensity of illumination |
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Answer» Correct Answer - A Kinetic energy of photoelectrons depends on the frequency of incident radiations and is independent of the intensity of illumination. |
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| 224. |
The magnitude of saturation photoelectric current depends uponA. FrequencyB. IntensityC. Work functionD. Stopping potential |
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Answer» Correct Answer - B The value of saturation current depends on intensity . It is independent of stopping potential |
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| 225. |
If the kinetic energy of a free electron doubles , its de - Broglie wavelength changes by the factorA. `(1)/(sqrt(2))`B. `sqrt(2)`C. `(1)/(2)`D. `2` |
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Answer» Correct Answer - A `lambda = (1)/(sqrt(2 mE)) rArr lambda prop (1)/(sqrt(2))`. |
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| 226. |
In a photoelectric effect , the `K.E.` of electrons emitted from the metal surface depends uponA. Intensity of lightB. Frequency of the incident lightC. Velocity of incident lightD. Both intensity and velocity of light |
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Answer» Correct Answer - B If frequency of incident light increases , kinetic energy of photoelectron also increases. |
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| 227. |
It is not possible for a photon to be completely absorbed by a free electron. Explain. |
| Answer» The free electron in metal is bound by some restraining forces. Due to which it can not come out of its own. Some energy is to be spent in ejecting the free electron form the surface of metal which is equal to work function of metal. It means, some of the energy of the incident photon is spent in liberting the electron form the surface of metal and remaining energy of incident photon is used in imparting the K.E. to the emitted electron. As the maximum K.E. of the emitted photoelectron is less than the energy of incident photon, hence it is not possible for a photon to be completely absorbed by a free electron. | |
| 228. |
A good quality mirror reflects about 75% of light incident on it. How will you find out, whether 25% of the photons have not been reflected at all or all the photons have been reflected but energy of each has been reduced by 25%? |
| Answer» First statement is true, because if energy of each photon were reduced, then the wavelength of reflected light would have been increased and the colour of reflected light would have changed. | |
| 229. |
For intensity `I` of a light of wavelength `5000 Å` the photoelectron saturation current is `0.40 mu A` and stopping potential is `1.36 V` , the work function of metal isA. `2.47 eV`B. `1.36 eV`C. `1.10 eV`D. `0.43 eV` |
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Answer» Correct Answer - C By using `E = W_(0) + K_(max)` `rArr E = (12375)/(5000) = 2.475 eV` and `K_(max) = eV_(0) = 1.36 eV` So `2.475 = W_(0) + 1.36 rArr W_(0) = 1.1 eV` |
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| 230. |
How does the maximum kinetic energy of electrons emitted very with work uniform of the metal? |
| Answer» As, `(K.E)_(max)=hv-phi_0`, so maximum K.E. of emitted electrons decreases with the increase in work function of the metal. | |
| 231. |
For intensity `I` of a light of wavelength `5000 Å` the photoelectron saturation current is `0.40 mu A` and stopping potential is `1.36 V` , the work function of metal is |
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Answer» Correct Answer - 1.1 eV `E=w_0=K_"max", E="hc"/lambda` |
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| 232. |
Light of frequency .25 `v_0` is incident on a metal surface of threshold frequency 2`v_0`. If its frequency is halved and intensity is made three times then find the new value of photoelectric current. |
| Answer» Zero, because the frequency of the incident light becomes less than the threshold frequency. | |
| 233. |
If the intensity of the incident light (whose frequency is greater than threshold frequency) is increased then explain its effect on (i) the maximum momentum that photoelectron could have and (ii) the minimum de -Broglie wavelength that photoelectron could have. |
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Answer» (i) Maximum K.E. of the emitted photo-electron of momentum `p_(max)` is given by `K_(max)=(p_(max)^2)/(2m)=hv-phi_0` `p_(max)=[2m(hv-phi_0)]^(1//2)` Where `phi_0` is the work function of the metal surface used on which light falls. The above relation is independent of intensity of the incident light (i.e., number of photons falling per second per unit area of light)but depends on the frequency of the incident light. If intensity of the incident light is increased without charging its frequency, the maximum momentum of the emitted photoelectrons remains unchanged. (ii) The minimum de-Broglie wavelength of the emitted photoelectron of maximum momentum `p_(max)` is, `lambda_(min)=h/(p_(max))` It is alos independent of intensity of incident light. Thus, minimum de-Broglie wavelength of emitted photoelectron remains unchanged with the increase in intesity of the incident light. |
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| 234. |
According to de - Broglie , the de - Broglie wavelength for electron in an orbit of hydrogen atom is `10^(-9) m`. The principle quantum number for this electron is |
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Answer» Correct Answer - 3 `mvr=(nh)/(2pi) rArr 2pir=nlambda` |
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| 235. |
In the nth orbit of hydrogen atom, find the ratio of the radius of the electron orbit and de-Broglie wavelength associated with it. |
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Answer» In nth orbit of hydrogen atom, the velocity `v_n` of electron is given by `mv_nr_n=(nh)/(2pi)` or `v_n=(nh)/(2pimr_n)` de-Broglie wavelength, `lambda=h/(mv_n)=h/mxx(2pimr_n)/(nh)=(2pir_n)/n` `:. (r_n)/lambda=n/(2pi)` |
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| 236. |
If m is the mass of an electron and c the speed of light, the ratio of the wavelength of a photon of energy E to that of the electron of the same energy isA. `csqrt((2m)/(E))`B. `sqrt((2m)/(E))`C. `sqrt((2m)/(cE))`D. `sqrt((m)/(E))` |
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Answer» Correct Answer - A (a) : energy of photon, `E=hv=(hc)/(lamda_(ph))` where `lamda_(ph)` is the wavelength of a photon, `lamda_(ph)=(hc)/(E)` Wavelength of the electron, `lamda_(e)=(h)/(sqrt(2mE))` `:.(lamda_(ph))/(lamda_(e))=(hc)/(E)xx(sqrt(2mE))/(h)=csqrt((2m)/(E))` |
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| 237. |
Choose the incorrect statement:A. The velocity of photoelectrons is directely proportional to the square root of wavelength of lightB. The number of photoelectrons emitted depeds upon the intensity of incident lightC. The velocity of photoelectron is directely proportional to the frequency of the incident lightD. The velocity of the photoelectrons is inversely proportional to square root of the frequency of the light. |
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Answer» Correct Answer - a,c,d As `K=hv-phi_(0)` or `1/2mv^(2)=(hc)/lambda-phi_(0)` `:. v prop sqrt(v)` or `v prop1/(sqrt(lambda))` so option `a,c,d` are incorrect as no. of photoelectrons depends on intensity of incident light, so, option b is correct. |
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| 238. |
What will be the ratio of de - Broglie wavelengths of proton and `alpha` - particle of same energy ? |
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Answer» Correct Answer - A `lambda = (h)/(sqrt(2 m E)) rArr lambda prop (1)/(sqrt(m)) rArr (lambda_(p))/(lambda_(p)) sqrt((m_(a))/(m_(p))) = (2)/(1)` |
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| 239. |
What is the de - Broglie wavelength of the alpha - particle accelerated through a potential difference `V`?A. `(0.287)/(sqrt(V)) Å`B. `(12.27)/(sqrt(V)) Å`C. `(0.101)/(sqrtV)Å`D. `(0.202)/(sqrt(V))Å` |
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Answer» Correct Answer - C `lambda = (h)/(sqrt( 2 m E)) = (h)/(sqrt(2 m_(alpha) Q_(alpha) V))` On putting `Q_(alpha ) = 2 xx 1.6 xx 10^(-19)` `m_(alpha) = 4m_(p) = 4 xx 1.67 xx 10^(-27) kg rArr lambda = (0.101)/(sqrt(V)) Å` |
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| 240. |
The speed of an electron having a wavelength of `10^(-10) m` isA. `7.25 xx 10^(6) m//s`B. `6.26 xx 10^(6) m//s`C. `5.25 xx 10^(6) m//s`D. `4.24 xx 10^(6) m//s` |
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Answer» Correct Answer - A By using `lambda_(el ectron) = (h)/( m_(e) v) rArr v = (h)/( m_(e) lambda_(e))` ` = (6.6 xx 10^(-34))/(9.1 xx 10^(-31) xx 10^(-10)) = 7.25 xx 10^(6) m//s`. |
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| 241. |
Compute the typical de-Broglie wavelength of an electrons in a metal at `27^(@)C` and compare it with the mean separation between two electrons in a metal which is given to be about `2xx10^(-10)m`. |
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Answer» Given , T `= 27 + 273 = 300 K, r = 2xx 10^(-10)` m Momemtum , P = ` sqrt(3" m KT") = sqrt(3xx 9.11 xx10^(31) xx1.38 xx10^(-23) xx300 ) = 1.06xx10^(-25)` kg -m/s `lambda = h/p = (6.63 xx10^(-34))/(1.06 xx10^(-25)) = 62.6 xx10^(-10) ` m , Mean seperation `r = 2xx 10^(-10)` m ` lambda/r = (62.6 xx10^(-10))/(2 xx10^(-10))= 31.3` We can see that de - Brogile wavelength is much greater than the electron seperation . |
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| 242. |
Light of wavelength 180 nm ejects photoelectrons from a plate of metal whose work - function is 2 eV. If a uniform magnetic field of `5xx10^(-5)` T be applied parallel to the plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy. |
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Answer» Max. K.E. of the ejected photoelectons is `K_(max)=1/2mv_(max)^(2)=(hc)/lambda-phi_(0)` `=((6.62xx10^(-34))xx(3xx10^(8)))/(180xx10^(-9))-2xx1.6xx10^(-10)` `=7.8xx10^(-19)J` `v_(max)=sqrt((2K_(max))/m)=sqrt((2xx7.8xx10^(-19))/(9.1xx10^(-31)))` `=1.3093xx10^(6)m//s` Radius of circular path in the magnetic field of induction B is given by , `Bev=mv^(2)//r` or `r=(mv)/(eB)=((9.1xx10^(-31))xx(1.3093xx10^(6)))/((1.6xx10^(-19))xx(5.0xx10^(-5)))` `=0.149m=14.9cm` |
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| 243. |
When a surface is irradiated with light of wavelength `4950Å`, a photocurrent appears which vanishes if a retarding potential appears which vanishes if a retarding potential greater than 1.2 volt is applied across the phototube. When a different source of light is used, it is found that the critical retarding potential is changed to 2.1 volt. Find the work function of the emitting surface and the wavelength of second source. If the photoelectrons (after emission from the surface) are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials. Use `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C`. |
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Answer» Case (i) `eV_(1)=(hc)/(lambda_(1))-phi_(0) or phi_(0)=(hc)/(lambda_(1))-eV_(1)` `:. phi_(0)=((6.6xx10^(-34))xx(3xx10^(8)))/(4950xx10^(-10))-(1.6xx10^(-19))xx1.2` `=2.08xx10^(-19)J=(2.08xx10^(-19))/(1.6xx10^(-19))eV=1.3eV` Case (ii) `eV_(2)=(hc)/(lambda_(2))-phi_(0) or (hc)/(lambda_(1))-eV_(2)+phi_(0)` `=(1.6xx10^(-19))xx1.2+2.08xx10^(-19)` `=5.44xx10^(-19)J` `lambda_(2)=(hc)/(5.44xx10^(-19)` `=((6.6xx10^(-34))xx(3xx10^(8)))/(5.44xx10^(-19))=3639.7Å` When photoelectrons are subjected to a perpendicular magnetic field, there will be no change in the retarding potential as the speed of the photoelectrons remains unchanged but the photoelectron simply describes a circular path in the field. |
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| 244. |
A small plate of a metal (work function `=1.7eV`) is placed at a distance of 2m from a monochromatic light source of wavelength `4.8xx10^(-7)m` and power 1.0watt. The light falls normally on the plate. Find the number of protons striking the metal plate per square metre per second. If a constant magnetic field of strength `10^(-4)T` is applied parallel to the metal surface, find the radius of the largest circular path followed by the emitted photoelectron. (use `h=6.63xx10^(-34)Js` , mass of electron=`9.1xx10^(-31)kg`, charge of electron=`1.6xx10^(-19)C` |
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Answer» Photon energy `=(hc)/lambda=((6.6xx10^(-34))xx(3xx10^(8)))/(4.8xx10^(-7))` `=4.125xx10^(-19)J` The rate of emission of photon from the source `=(1.0Js^(-1))/(4.125xx10^(-19)J)=2.425xx10^(18)s^(-1)` No. of photon striking per square meter per second on the metal plate `=(2.425xx10^(18))/(4pir^(2))=(2.425xx10^(18))/(4xx(3.14)xx(2)^(2))=4.82xx10^(16)` The maximum K.E. (K) of the photoelectrons emitted from the plate having work function `phi_(0)(=1.17eV)` is given by `K_(max)=(hc)/lambda-phi_(0)` `=(4.125xx10^(-19)J)-(1.17xx1.6xx10^(-19)J)` `=2.253xx10^(-19)J` The maximum velocity of photoelectrons `v_(max)=sqrt((2K_(max))/m)=sqrt((2xx2.253xx10^(-19))/(9.1xx10^(-31)))` `=7.036xx10^(5)ms^(-1)` The radius of the largest circle traversed by photoelectron in magnetic field B in given by `r_(max)=(mv_(max))/(eB)=((9.1xx10^(-31))xx(7.036xx10^(5)))/((1.6xx10^(-19))xx(10^(-4)))` `=0.04m=4.0cm` |
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| 245. |
For characteristic `X` - ray of some materialA. `E(K_(gamma)) lt E(K_(beta)) lt E(K_(alpha))`B. `E(K_(alpha)) lt E(L_(alpha)) lt E(M_(alpha))`C. `lambda(K_(gamma)) lt lambda(K_(beta)) lt lambda(K_(alpha))`D. `lambda(M_(alpha)) lt lambda(L_(alpha)) lt lambda(K_(alpha))` |
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Answer» Correct Answer - C `because E(K_(gamma)) gt E(K_(beta)) gt E (K_(alpha)) rArr lambda (K_(gamma)) lt lambda (K_(beta)) lt lambda (K_(alpha))` |
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| 246. |
Monochromatic light of wavelength `3000 Å` is incident on a surface area `4 cm^(2)`. If intensity of light is `150 m W//m^(2)` , then rate at which photons strike the target isA. `3 xx 10^(10) //sec`B. `9 xx 10^(13)//sec`C. `7 xx 10^(15)//sec`D. `6 xx 10^(19)//sec` |
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Answer» Correct Answer - B `(n)/(t) = (I A lambda)/(hc) = (150 xx 10^(-3) xx 4 xx 10^(-4) xx 3 xx 10^(-7))/( 6.6 xx 10^(-34) xx 3 xx 10^(8))` `= 9 xx 10^(13) (1)/( sec)` |
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| 247. |
Find the maximum velocity of photoelectrons emitted by radiation of frequency `3 xx 10^(15)` Hz from a photoelectric surface having a work function of 4.0 eV. Given `h=6.6 xx 10^(-34)` Js and 1 eV = `1.6 xx 10^(-19)` J. |
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Answer» `1/2mv_"max"^2=hv-phi` `=6.63xx10^(-34)xx3xx10^15 - 4xx1.6xx10^(-19)` `v_"max"^2=(2[19.89xx10^(-19)-6.4xx10^(-19)])/(9.1xx10^(-31))` `=(26.98xx10^(-19))/(9.1xx10^(-31))` `=2.96xx10^12` `v_"max"=1.72xx10^6` m/s |
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| 248. |
A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of `10fm` to the nucleus. The de - Broglie wavelength (in units of fm) of the proton at its start is take the proton mass, `m_p = 5//3xx10^(-27) kg, h//e = 4.2xx10^(-15) J-s//C`, `(1)/(4piepsilon_0) = 9xx10^9m //F, 1 fm = 10^(-15)`.A. 7B. 9C. 11D. 13 |
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Answer» Correct Answer - A (a) : As `(1)/(4piepsilon_(0))((120e)(e))/(10xx10^(-15))=(P^(2))/(2m_(p))` . . .(i) where p is the momentum of the proton and `m_(p)` is the mass of the proton de Broglie wavelength of proton, `lamda=(h)/(p)orp=(h)/(lamda)` Substituting this value of p in equation (i), we get `(1)/(4piepsilon_(0))(120e^(2))/(10xx10^(-15))=(h^(2))/(2lamda^(2)m_(p))` `lamda^(2)=(4piepsilon_(0)xx10xx10^(-15)xxh^(2))/(2m_(p)xx120e^(2))` Substituting the given numerical values, we get `lamda^(2)=(1xx10xx10xx^(-15)xx4.2xx10^(-15)xx4.2xx10^(-15))/(9xx10^(9)xx2xx(5)/(3)xx10^(-27)xx120)` `(42xx42xx10^(-30)xx10^(-14))/(36xx10^(-14))=49xx10^(-30)` `lamda=7xx10^(-15)m=7fm` |
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| 249. |
Light of wavelength 2000 Å falls on a metallic surface whose work function is 4.21 eV. Calculate the stopping potential |
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Answer» `eV_s="hc"/lambda-phi_0` or `V_s="hc"/(elambda)-phi_0/e` `V_s=(6.62xx10^(-34)xx3xx10^8)/(1.6xx10^(-19)xx2000xx10^(-10))-4.21` or `V_s`=6.21-4.21= 2 V |
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| 250. |
if the speed of photoelectrons is `10^4` m/s, what should be the frequency of the incident radiation on a potassium metal ? Given : Work function of potassium =2.3 eV |
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Answer» `E=K_"max"+phi` `=1/2mv^2+phi` `=1/2xx9xx10^(-31)xx(10^4)^2+2.3xx1.6xx10^(-19)` `=4.5xx10^(-23) J + 3.68 xx 10^(-19)` `=3.68xx10^(-19)` J `v=E/h=(3.68xx10^(-19))/(6.63xx10^(-34))` Hz `=0.56xx10^15` Hz |
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