InterviewSolution
Saved Bookmarks
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.
| 51. |
Assertion : The stopping potential depends on the frequency of incident light. Reason : The stopping potential is related to maximum kinetic energy by `eV_(0)=K_(max)`.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 both assertion and reason are false. |
|
Answer» Correct Answer - A (a) : The stopping potential is related to the maximum kinetic energy of electrons emitted as `eV_(0)=(1)/(2)mv_(max)^(2)=K_(max)` Now, `K_(max)=hv-phi_(0)`. Therefore, stopping potential `V_(0)` depends on the frequency of incident light. |
|
| 52. |
fig. shows variation of stopping potential `(V_0)` with frequency (v) for two photosenstive matrials `M_1 and M_2`. (i) why is the slope same for both lines? (ii) For which material will the emitted electrons have greater kinetic energy for the incident radiations of the same frequency? Justify your answer. |
|
Answer» (i) From photoelectric equation, we have `eV_0-hv-phi_0 or V_0=(hv)/e-(phi_0)/e` It is an equation of straight line with slope h/e (=constant). It means the slope metals `V_0-v` graph (=h/e) is same for both metals `M_1 and M_2`. (ii) Also, `K_(max)=hv-hv_0`. For the given frequency of incident light, the smaller is the value of `v_0`, the larger is the value of `K_(max)` and vice versa. Since material `M_1` has lower value of threshold frequency `v_0`, so metal `M_1` will emit photoelectrons of greater K.E. |
|
| 53. |
What is the de-Broglie wavelength associated with an electron moving with an electron, accelerated through a potential difference of 100 volt? |
|
Answer» Accelerating potential V = 100 V. The de Broglie wavelength `lambda` is `lambda=h//p=1.227/sqrtV nm` `lambda=1.227/sqrt100 nm =0.123 nm ` The de Broglie wavelength associated with an electron in this case is of N8the order of X-ray wavelengths. |
|
| 54. |
What is the threshold wavelength for a cesium surface, for which the work function is 1.8 eV ? |
|
Answer» Correct Answer - 689 nm 689 nm |
|
| 55. |
Sketch a graph between frequency of incident radiations and stopping potential for a given photosenstive material. What information can be obtained from the value of the intercept on the potential axis? A source of light of frequency greater than the threshold frequency is placed at a distance of 1m from the cathode of a photocell. The stopping potential is found to be V. If the distance of the light source from the cathode is reduced, explain giving reasons, what change will you observe in the (i) photoelectric current (i) stopping potential? |
| Answer» For graph intercept on the potential axis=`-phi_0//e`. If the distance of the light source of the cathode is reduced, then (i) the photoelectric current is increases because the intensity of incident radiation increases. (ii) the stopping potential `V_0` remains the same because it is independent of the intensity of the incident radiations. | |
| 56. |
The slope of frequency of incident light and stopping potential for a given surface will beA. `h`B. `h//e`C. `eh`D. `e` |
| Answer» Correct Answer - b | |
| 57. |
The following figure shows a graph for the stopping potential as a function of the frequency of incident light illuminating a metal surface. Find the photoelectric work function for this surface. |
|
Answer» Correct Answer - 2.0 eV 2.0 eV |
|
| 58. |
The speed of an electron having a wavelength of `10^(-10) m` is |
|
Answer» Correct Answer - `7.25xx10^6` m/s `lambda=h/"mv"` |
|
| 59. |
If the intensity of the incident radiation in a photocell is increased, how does the stopping potential very? |
| Answer» As, `eV_0=h(v-v_0)`, so stopping potential `V_0` remains unchanged by the increase in intensity of the incident radiation. | |
| 60. |
Why is a photoelectric cell also called an electric eye?A. LEDB. PhotocellC. Integrated chip (IC)D. Solar cell |
| Answer» Correct Answer - B | |
| 61. |
In Davisson and Germer experiment, the tungsten filament is coated withA. aluminium oxideB. barium chlorideC. titanium oxideD. barium oxide |
| Answer» Correct Answer - D | |
| 62. |
The energy of a photon is `E = hv` and the momentum of photon `p = (h)/(lambda)` , then the velocity of photon will beA. `E//p`B. `Ep`C. `((E )/(P))^(2)`D. `3 xx 10^(8) m//s` |
|
Answer» Correct Answer - A Momentum of photon `p = (E )/( c )` `rArr` Velocity of photon ` c = ( E)/( p)` |
|
| 63. |
An electron, an `alpha`-particle and a proton have the same kinetic energy. Which of these particles has the largest de-Broglie wavelength? |
|
Answer» Kinetic energy `K=1/2 mv^2=1/2 (m^2v^2)/m=1/2 (p^2)/m` or `p=sqrt(2mK)` De-Broglie wavelength, ` lambda=h/p=h/(sqrt(2mK))` `:. lambda prop1/(sqrtm)` (where K is constant) It means smallest is the particle mass, largest is the de-Broglie wavelength associated with it. Since the mass of electron is least out of the given particles, hence d-Broglie wavelength is largest for electron and least for `alpha`-particle. |
|
| 64. |
Find the ratio of de Broglie wavelength of an `alpha` -particle and a deutron if they are accelerating through the same potential difference |
|
Answer» Correct Answer - `1/2` `lambda=h/sqrt(2mqV)` |
|
| 65. |
Assertiion : `X` - rays are used for studying the structure of crystals. Reason : The distance between the atoms of crystals is of the order of wavelength of `X` - 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. |
|
Answer» Correct Answer - A The distancce between the atoms of crystals is of the order of wavelength of `X` - rays . When they fall on a crystal , they are diffracted. The diffraction pattern is helpful in the study of crystal structure. |
|
| 66. |
In which of the following photocell is not used ?A. Burglar alarmB. Television cameraC. Automatic street lightsD. Vacuum cleaner |
| Answer» Correct Answer - D | |
| 67. |
Light of intensity `10^(-5)Wm^(-2)` falls on a sodium photocell of surface area `2cm^(2)`. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave picture of radiation. The work function of the metal is given to be about 2eV. What is the implication of your answer? effective atomic area `=10^(-20)m^(2)`. |
|
Answer» Given , I = ` 10^(5) W//m^(2) , A = 2 cm ^(2) = 2 xx 10^(-4) " m"^(2) , phi _(0) = 2eV ` Let t be the time The effective atomic area of Na = `0^(-20) m^(2)` nand it contains conduction electron per atom . No. of conduction electrons in five layers `= (5 xx "Area of one laye")/("Effective atomic area ") = (5xx 2 xx 10^(-4))/(10^(-20) = 10^(17)` We know that sodium has one free electron (or conduction electron ) per atom . Incident power on the surface area of photocell = Incident intensity `xx` Area on the surface area of photo cell ` 10^(-5) xx 2xx 10^(-4) = 2 xx 10^(-9) ` W The electron present in all the 5 layers of sodium will share the incident energy equally Energy absorbed per second per electron , E = ` ("Incident power ")/(" No. of electrons in 5 layers ")` ` = ( 2 xx 10^(-9))/(10^(17)) = 2 xx 10^(-26)` W Time required for mission by each electron , `T = ("Energy required per electron ")/("Energy absorbed per second ") = (2 xx 1.6 xx 10^(-19))/(2 xx10^(-26)) = 1.6 xx10^(7) ` s , which is about 0.5 yr. The answer obtained implies that the tme of emission of electron is very large and is not agreement with the observed time of emission of photoelectron . Thus, it is implied that the wave theory cannot be applied in this experiment . |
|
| 68. |
Statement-1:An electron cannot be observed without changing its momentum by an indeterminate amount. Statement-2:Momentum of an electron is `h/lambda` where `lambda` is its wavelengthA. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - 2 | |
| 69. |
Calculate the ratio of the accelerating potential required to accelerate (i) a proton and (ii) and `alpha` particle to have the same de-Broglie wavelength associated with them. |
|
Answer» If the charged particle of mass m, charge q is accelerated under a pot. diff. V, the velocity acquired by particle is v. Then `qV=1/2mv^2=(m^2v^2)/(2m)=(p^2)/(2m)` Where p is momentum of particle of `p=sqrt(2mqV)` De-Broglie wavelength, `lambda=h/p=h/(sqrt(2mqV))` Now, `lambda_p=h/(sqrt(2m_pq_pV_p)) and lambda_(alpha)=h/(sqrt(2m_(alpha)q_(alpha)V_(alpha)))` of `(V_p)/(V_(alpha))=(m_(alpha)q_(alpha))/(m_pq_p)=(4mxx2e)/(mxxe)=8` |
|
| 70. |
Assertion: Photoecell is called electric eye. Reason: Photocell can see tha things placed before it.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
|
Answer» Correct Answer - c Here Assertion is correct but Reason is wrong. Photocell converts light energy into electric current energy which is used for velocity of uses. |
|
| 71. |
Assertion: The de-broglie wavelength equation has significance for any microscopic or submicroscopic particles. Reason : The de-broglie wavelength is inversely proportional to the mass of the object if velocity is constant.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
|
Answer» Correct Answer - a Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion. |
|
| 72. |
Statement-1:An electron microscope can achieve better resolving power than an optical microscope Statement-2:The de Broglie wavelength of the electrons emitted from an electron gun is much less than 500 nmA. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-2B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - 1 | |
| 73. |
Electrons used in an electron microscope are accelerated by a voltage of `25 kV`. If the voltage is increased to `100 kV` then the de - Broglie wavelength associated with the electrons wouldA. decrease by `2` timesB. decrease by `4` timesC. increase by `4` timesD. increase by `2` times |
|
Answer» Correct Answer - A We have `lambda = (12.27)/(sqrt(V))` So , `(lambda_(1))/(lambda_(2)) = sqrt((V_(2))/(V_(1))) rArr lambda_(2) = lambda_(1) sqrt((V_(1))/(V_(2)))` `lambda_(2) = lambda_(1) sqrt((25)/(100)) rArr lambda_(2) = lambda_(1) sqrt((1)/(2)) = (lambda_(1))/(2)` |
|
| 74. |
Assertion: An electron microscop can achieve better resolving power then an optical microscope. Reason : The de-broglie wavelength of the electrons emitted form an electron gun with velocity `500m//s` is much less than `500nm`.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
|
Answer» Correct Answer - b Both Assertion and Reason are correct but the Reason is not correct explanation of the Assertion. |
|
| 75. |
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. The saturation photocurrent gets doubleB. The saturation photocurrent remains almost the sameC. The maximum K.E. of the photoelectron is more than doubledD. The maximum K.E. of the photoelectrons gets double. |
|
Answer» Correct Answer - a,c Saturation photoelectric current depends upon the intensity of the incident light. As `K_(max)=hv-phi_(0)`. If `v` becomes double, then `K_(max)` becomes more than double as `2hv-phi_(0)gt2(hv-phi_(0))` |
|
| 76. |
Solid targets of different elements are bombarded by highly energetic electron beam. The frequcny `(f)` of the characteristic `X`-rays emitted from different targets varies with atomic number `Z` asA. `f prop sqrt(Z)`B. `f prop Z^(2)`C. `f prop Z`D. `f prop Z^(3//2)` |
|
Answer» Correct Answer - B `lambda prop (1)/(Z^(2)) rArr ( c )/(v) prop (1)/(Z^(2)) rArr v prop Z^(2)` |
|
| 77. |
(a) Obtain the de-Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in previous problem 31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. Given `m_(n)=1.675xx10^(-27)kg`. (b) Obtain the de-Broglie wavelength associated with thermal neutrons at room temperature `(27^(@)C)`. Hence explain why a fast neutrons beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments. |
|
Answer» De Broglie wavelength `=2.327xx10^(-12)m`, neutron is not suitable for the diffraction exeriment Kinetic energy of the neutron, `K=150eV =150xx1.6xx10^(-19)` `2.4xx10^(-17)J` Mass of a neutron, `M_(n)=1.675xx10^(-27)Kg` The kinetic energy of the neutron is given by the relatin : `K=(1)/(2)m_(n)v^(2)` `m_(n)v=sqrt(2Km_(n))` Where, v=Velocity of the neutron `m_(n)v`=Momentum of the neutron De-Broglie wavelength of the neutron is given as : `lambda =(h)/(m_(n)v)=(h)/sqrt(2Km_(n))` It is clear that wavelength is inversely proportional to the square root of mass. Hence, wavelength decreases with increase is mass and vice versa. `therefore lambda =(6.6xx10^(-34))/sqrt(2xx2.4xx10^(-17)xx1.675xx10^(-27))` =`2.327xx10^(-12)m` It is given in the previous problem that the inter-atomic spacing of a crystal is about `1Å`, i.e., `10^(-10)`m.Hence, the inter- atomic spacing is about a hundred times greater, Hence a neutron beam of energy 150 eV is not suitable for diffraction experiments. (b) De Broglie wavelegth =`1.447xx10^(-10)m` Room temperature ,`T=27^(@)C=27+273=300K` The average kinetic energy of the neutron is given as : `E=(3)/(2)kT` Where, K= Boltzmann constant `=1.38xx10^(-23)JMol^(-1)K^(-1)` The wavelength of the neutron is given as : `lambda=(h)/sqrt(2M_(n)E)=(h)/sqrt(3M_(n)kT)` `=(6.6xx10^(-34))/sqrt(3xx1.675xx10^(-27)xx1.38xx10^(-23)xx300)` `=1.447xx10^(-10)m` This wavelength is comparable to the inter-atmomic spacing of a crystal. Hence, the high-energy neutron beam should first be thermalised, before using it for diffraction. |
|
| 78. |
Obtain the de Broglie wavelength associated with thermal neutrons at room temperature `(27^(@)C)` . Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments . |
|
Answer» `T = t+273 = 27 +273 = 300 K, K = 1.38 xx 10^(-23)` J /mol/K ` E = 3/2 KT rArr lambda = h/(2"mE") = h/(sqrt(2mxx3/2 KT)) = h/(3"mKT")` ` (6.63 xx10^(-34))/(sqrt(3 xx 1.675 xx10^(-27) xx 1.38 xx10^(-23) xx 300 )) = 1.45 xx10^(-10)` m This wavelength is order of interatomic spacing So, the neutron beam first thermalised and then used for diffraction . |
|
| 79. |
For what kinetic energy of a neutron will the associated de-Broglie wavelength be `1.40 xx 10^(-10)m` ? Mass of neutron = `1.675 xx 10^(-27) kg, h = 6.63 xx 10^(-34) Js`. |
|
Answer» Givn, for neutron, `lambda = 1.40xx10^(-10)m` and `m = 1.675xx10^(-27)kg` `KE = (P^(2))/(2m)=(h^(2))/(2m lambda^(2))= ((6.63xx10^(-34))^(2))/(2xx(1.40xx10^(-10))^(2)xx1.675xx10^(-27))therefore KE = 6.686xx10^(-21)J` |
|
| 80. |
A photon , an electron and a uranium nucleus all have the same wavelength . The one with the most energyA. is the photonB. is the electronC. is the uranium nucleusD. depends upon the wavelength and the properties of the particle |
|
Answer» Correct Answer - A `lambda = (h)/( mv) = (h)/(sqrt(2 m E)) :. E = (h^(2))/( 2m lambda^(2))` `lambda` is same for all , so `E prop (1)/(m)`. Hence energy will be maximum for particle with lesser mass. |
|
| 81. |
The de - Broglie wavelength `lambda` associated with an electron having kinetic energy `E` is given by the expressionA. `(h)/(sqrt( 2 m E))`B. `( 2h)/(m E)`C. `2 mhE`D. `( 2 sqrt(2 mE))/(h)` |
|
Answer» Correct Answer - A `(1)/(2) mv^(2) = E rArr = mv = sqrt(2 m E) , :. E = (h)/( mv) = (h)/(sqrt( 2m E))` |
|
| 82. |
The log - log graph between the energy `E` of an electron and its de - Broglie wavelength `lambda` will beA. B. C. D. |
|
Answer» Correct Answer - C `l = (h)/(sqrt( 2 m E)) = (1)/(sqrt(2 m E)) . (1)/(sqrt ( E))`. Taking log of both sides `log lambda = log (1)/(sqrt (2 m)) + log (1)/( sqrt (E ))` `rArr log lambda = log (h)/(sqrt ( 2m)) - (1)/(2) log E` `rArr log lambda = -(1)/(2) log E + log (h)/(sqrt( 2 m))` This is the equation of straight line having slope `( - 1//2)` and positive intercept on `log lambda` axis. |
|
| 83. |
Calculate the Momentum, |
|
Answer» Given, `V = 56 V, e = 1.6xx10^(-19)c, m = 9xx10^(-31)kg` As `KE=(P^(2))/(2m) rArr 2m(KE)= P^(2) rArr P = sqrt(2m(KE)) = sqrt(2m eV) [because KE = eV]` `therefore P = sqrt(2xx9xx10^(-31)xx1.6xx10^(-31)xx56) = 4.02xx10^(-24) kg-m//s` |
|
| 84. |
A photon in motion has a mass |
| Answer» Correct Answer - `m="hv"/c^2` | |
| 85. |
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. |
|
Answer» Given, `V = 56 V, e = 1.6xx10^(-19)c, m = 9xx10^(-31)kg` `lambda = (12.27)/(sqrt(V)) Å = (12.37)/(sqrt(56))= 0.164xx10^(-9)m therefore lambda = 0.164 nm`. |
|
| 86. |
A proton and an `alpha`-particle are accelerated through same potential difference. Find the ratio of their de-Brogile wavelength. |
|
Answer» `lambda=h/"mv"=h/sqrt(2mE)=h/sqrt(2mqV)[because E=qV]` For proton `m_p=m,q=e` For `alpha`-particle `m_alpha`=4 m, q=2 e `lambda_alpha/lambda_p=sqrt((m_pq_p)/(m_alphaq_alpha))` `=1/(2sqrt2)` |
|
| 87. |
Calculate the (a)momentum and (b) de - Brogile wavelength of the electrons accelerated through a potential difference of 56 V. |
|
Answer» Mass of the electron , m = `9 xx 10^(-31) " kg , "` Potential difference , V - 56 V Momentum of electrons , mv = ` sqrt ( 2 e V m ) = sqrt(2 xx (1.6 xx 10^(-19) )xx56 xx 9 xx 10^(-31))= 4.02 xx 10^(-24) " kg ms"^(-1)` (b) de - Broglie wavelength , ` lambda = h/p = h/(m v)= (6.62 xx 10^(-34))/(4.04 xx 10^(-24))= 1.64 xx 10^(-10) m ` |
|
| 88. |
The number of photons of wavelength `540 nm` emitted per second by an electric bulb of power `100 W` is (taking `h = 6 xx 10^(-34) sec`)A. `100`B. `1000`C. `3 xx 10^(20)`D. `3 xx 10^(18)` |
|
Answer» Correct Answer - C `P = (nhc)/(lambda t) rArr 100 = (n xx 6 xx 10^(-34) xx 3 xx 10^(8))/(540 xx 10^(-9) xx 1)` `rArr n = 3 xx 10^(20)` |
|
| 89. |
In a photoelectric experiment, if both the intensity and frequency of the incident light are doubled, then the saturation photoelectric currentA. remains constantB. is halvedC. is doubledD. becomes four times |
|
Answer» Correct Answer - C ( c) : The saturation photoelectric current is directly proportional to the intensity of incident is radiation but it is independent of its frequency. Therefore, the saturation photoelectric current becomes doubled, when both intensity and frequency of the incident light are doubled. |
|
| 90. |
Wavelength of a `1 ke V` photon is `1.24 xx 10^(-9) m`. What is the frequency of ` 1 Me V` photon ?A. `1.24xx10^15` HzB. `2.4xx10^20` HzC. `1.24xx10^18` HzD. `2.4xx10^23` hz |
| Answer» Correct Answer - 2 | |
| 91. |
Wavelength of a `1 ke V` photon is `1.24 xx 10^(-9) m`. What is the frequency of ` 1 Me V` photon ? |
|
Answer» Correct Answer - `3.3xx10^(-29)` kg m/s E=hv |
|
| 92. |
Which of the following statements is not correct ?A. Photographic plates are sensitive to infrared raysB. Photographic plates are sensitive to ultraviolet raysC. Infrared rays are invisible but can cast shadows like visible lightD. Infrared photons have more energy than photons of visible light |
| Answer» Correct Answer - 4 | |
| 93. |
If we express the energy of a photon in `KeV` and the wavelength in angstroms , then energy of a photon can be calculated from the relation |
|
Answer» Correct Answer - `E_"(keV)"=12.4/lambda` `E="hc"/lambda` Joules |
|
| 94. |
If we express the energy of a photon in `KeV` and the wavelength in angstroms , then energy of a photon can be calculated from the relationA. E=12.4 hvB. `E="12.4 h"/lambda`C. `E=12.4/lambda`D. E=hv |
| Answer» Correct Answer - 3 | |
| 95. |
If we express the energy of a photon in `KeV` and the wavelength in angstroms , then energy of a photon can be calculated from the relationA. `E = 12.4 hv`B. `E = 12.4 h//lambda`C. `E = 12.4// lambda`D. `E = hv` |
|
Answer» Correct Answer - C Energy of photon `E = (hc)/( lambda) (joules) = (hc)/( e lambda) ( eV)` `rArr E( eV) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx lambda(Å)) = (12375)/(lambda (Å))` `rArr E(keV) = (12.37)/( lambda (Å)) ~~ (12.4)/(lambda)` |
|
| 96. |
Statement-1: A small metal ball is suspended in a uniform electric field with an insulated thread. If high energy X-rays beam falls on the ball, the ball will be declected in the electric field. Statement-2: X-rays emit photoelectrons and metal and metal becomes negatively charged.A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
|
Answer» Correct Answer - c When high energy X-rays fall on the ball, the metal ball will emit photoelectrons, leaving the positive charge on the ball. As a result of it, the ball is deflected in the direction of the electric field. Thus, the statement-1, is true but statement-2 is true. |
|
| 97. |
In an `X` - rays tube , the intensity of the emitted `X` - rays beam is increased byA. Increasing the filament currentB. Decreasing the filament currentC. Increasing the target potentialD. Decreasing the target potential |
|
Answer» Correct Answer - A By changing the filament current with the help of rheostat , thermionic emission intensity of `X - rays` can be changed. |
|
| 98. |
Light of wavelength `4000 Å` falls on a photosensitive metal and a negative `2 V` potential stops the emitted electrons. The work function of the material ( in eV) is approximately `( h = 6.6 xx 10^(-34) Js , e = 1.6 xx 10^(-19) C , c = 3 xx 10^(8) ms^(-1))`A. `1.1`B. `2.0`C. `2.2`D. `3.1` |
|
Answer» Correct Answer - A Energy of incident light `E( eV) = (12375)/(4000) = 3.09 eV` Stopping potential is `- 2V` so `K_(max) = 2 eV` Hence by using `E = W_(0) + K_(max) rArr W_(0) = 1.09 ~~ 1.1` eV |
|
| 99. |
If the wavelength of light is `4000 Å`, then the number of waves in `1 mm` length will beA. `25`B. `0.25`C. `2.5`D. `25 xx 10^(4)` |
|
Answer» Correct Answer - C Number of waves `= (10^(-3))/( 4000 xx 10^(-10)) = 2.25 xx 10^(4)` |
|
| 100. |
If in a photoelectric experiment , the wavelength of incident radiation is reduced from `6000 Å to 4000 Å` thenA. Stopping potential will decreaseB. Stopping potential will increaseC. Kinetic energy of emitted electrons will decreaseD. The valuer of work function will decrease |
|
Answer» Correct Answer - B Stopping potential `V_(0) = (hc )/€ [ (1)/(lambda) - (1)/(lambda_(0))]`. As `lambda` decreases so `V_(0)` increases. |
|