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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.
| 101. |
Five elements `A , B , C , D` and `E` have work functions `1.2 eV , 2.4 eV , 3.6 eV , 4.8 eV` and `6 eV` respectively . If light of wavelength `4000 Å` is allowed to fall on these elements , then photoelectrons are emitted byA. `A , B` and `C`B. `A , B , C , D` and `E`C. `A` and `B`D. Only `E` |
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Answer» Correct Answer - C `E = (12375)/(4000) = 3.09 eV` Photoelectrons emits if energy of incident light gt work function. |
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| 102. |
Sodium and copper have work functions `2.3 eV and 4.5 eV` respectively . Then the ratio of the wavelength is nearestA. `1 : 2`B. `4 : 1`0C. `2 : 1`D. `1 : 4` |
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Answer» Correct Answer - C `W_(0) prop (1)/(lambda) rArr (lambda_(1))/( lambda_(2)) = ((W_(0))_(2))/((W_(0))_(1)) = (4.5)/(2.3) = (2)/(1)` |
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| 103. |
Photons absorbed in matter are converted to heat. A source emitting n photons/ sec of frequency `nu` is used to convert 1kg of ice at `0^(@)C` to water at `0^(@)C`. Then, the time T taken for the conversionA. decreases with increasing n, with v fixedB. decreases with n fixed, v increasingC. remains constant with n and b changing such that nv= constantD. increases when the product nv increases |
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Answer» Energy spent to convert ice into water `" "` = mass x laten heat `" "=mL=(1000g)xx(80cal//g)` `" "` = 80000 cal Energy of photons used `=nTxxE=nTxxhv` `So, " "nThv=mLimpliesT=(mL)/(nhv)" "[becauseE=hv]` `therefore " "Tprop1/n,` when n is fixed. `therefore " "Tprop1/v,` when n is fixed. `impliesTprop1/(nv)` Thus, T is constant, if nv is constant. |
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| 104. |
Photons absorbed in matter are converted to heat. A source emitting n photons/ sec of frequency `nu` is used to convert 1kg of ice at `0^(@)C` to water at `0^(@)C`. Then, the time T taken for the conversionA. decreases with increasing n, with v fixed.B. decreases with n fixed, v increasing.C. remains constant with n and v changing such that nv=constant.D. All of these. |
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Answer» Correct Answer - D (d) : Energy spent to convert ice into water `=mL=(1000g)xx80calg^(-1)=80000cal` Energy of photons used `=nTxxE=nTxxhv` `:.nThv=mL or T=(mL)/(mhv)` `:.Tprop1//n"when v is constant",Tprop1//v"when n fixed",Tprop1//nv."Thus T is constant if is constant. Thus options (a), (b) and (c) are correct"`. |
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| 105. |
A source `S_(1)` is producing `10^(15)` photons per second of wavelength `5000Ã…`. Another source `S_(2)` is producing `1.02xx10^(15)` Then, `("Power of" S_(2))//("Power of"S_(1))` is equal toA. `1.00`B. 1.02C. 1.04D. 0.98 |
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Answer» Correct Answer - A (a) : For a source `S_(1)`, Wavelength, `lamda_(1)=5000Ã…` Number of photons emitted per second, `N_(1)=10^(15)` Energy of each photon, `E_(1)=(hc)/(lamda_(1))` Power of source `S_(1),P_(1)=E_(1)N_(1)=(N_(1)hc)/(lamda_(1))` For a source `S_(2)`, Wavelength, `lamda_(2)=5100Ã…` Number of photons emitted per second, `N_(2)=1.02xx10^(15)` Energy of each photon, `E_(2)=(hc)/(lamda_(2))` Power of source `S_(2),P_(2)=N_(2)E_(2)=(N_(2)hc)/(lamda_(2))` `:.("Power of "S_(2))/("Power of "S_(1))=(P_(2))/(P_(1))=(lamda_(2))/((N_(1)hc)/lamda_(1))=(N_(2)lamda_(1))/(N_(1)lamda_(2))` `=((1.02xx10^(15)"photons"//s)xx(5000xx10^(-10)))/((10^(15)"photons"//s)xx(5100xx10^(-10)))=(51)/(51)=1` |
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| 106. |
A source `S_(1)` is producing `10^(15)` photons//s of wavelength `5000 Å` Another source `S_(2)` is producing `1.02 xx 10^(15)` photons per second of wavelength `5100 Å`. Then `("power of" `S_(2))//("power of" S_(1))` is equal toA. `1.00`B. `1.02`C. `1.04`D. `0.98` |
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Answer» Correct Answer - A Number of photons emitted per second ` n = (p)/(((hc)/(lambda)))` `:. P = ( nhc)/(lambda)` `rArr (P_(2))/(P_(1)) = (n_(2) lambda_(1))/(n_(1) lambda_(2)) = (1.02 xx 10^(15) xx 5000)/(10^(15) xx 5100) = 1` |
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| 107. |
How many photons are emitted by a laser source of `5 xx 10^(-3)` W operating at 632.2 nm in 2 second `(h = 6.63 xx 10^(-34)Js)`?A. `3.2xx10^16`B. `1.6xx10^16`C. `4xx10^16`D. `0.4xx10^16` |
| Answer» Correct Answer - 1 | |
| 108. |
The most proable kinetic energy of thermal neutrons at a temperature of T Kelvin, may be taken equal to kT, where k is Boltzmann constant. Taking the mass of a neutron and its associated de-Broglie wavelength as m and `lambda_B` respectively, state the dependence of `lambda_B` on m and t. |
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Answer» De-Broglie wavelength, `lambda=h/p=h/(sqrt(2mK))=h/(sqrt(2mxx(3/2kT)))` `:. lambda_B prop 1/(sqrt(mT))` |
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| 109. |
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 `0.2 V`B. the stopping potential will be `0.6 V`C. the saturation current will be `6 mA`D. the saturation current will be `18 mA` |
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Answer» Correct Answer - C Cut off `"voltage"` is independent of intensity and hence remains the same. Since distance becomes `3 times`, so intensity `(I)` becomes `(I)/(9)`. Hence photo current also decreases by this factor i.e., becomes `(18)/(9) = 2 mA`. |
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| 110. |
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. Decrease by a factor of `2`B. Increase by a factor of `2`C. Decrease by a factor of `4`D. Increase by a factor of `4` |
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Answer» Correct Answer - D Number of photoelectrons `prop (1)/(( Distance)^(2))`. |
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| 111. |
A radio transmitter operates at a frequency of `880 kHz` and a power of `10 kW`. The number of photons emitted per second areA. `1.72 xx 10^(31)`B. `1327 xx 10^(34)`C. `13.27 xx 10^(34)`D. `0.075 xx 10^(-34)` |
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Answer» Correct Answer - A Number of photons emitted per second `n = (p)/( hv) = (10 xx 10^(3))/( 6.6 xx 10^(-34) xx 880 xx 10^(3)) = 1.72 xx 10^(31)` |
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| 112. |
A radio transmitter operates at a frequency of `880 kHz` and a power of `10 kW`. The number of photons emitted per second are |
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Answer» Correct Answer - `1.72xx10^(31)//sec` `n=P/(hv)=(10xx1000)/(6.6xx10^(-34)xx(880xx1000))` `1.72xx10^(31)//sec` |
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| 113. |
Assertion : The wave nature of electrons was first experimentally verified by Davisson and Germer Experiment. Reason : From the electron diffraction measurements, the wavelength of metter waves was found to 0.165 nm.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 - B | |
| 114. |
An electron of mass `m` when accelerated through a potential difference `V` has de - Broglie wavelength `lambda`. The de - Broglie wavelength associated with a proton of mass `M` accelerated through the same potential difference will beA. `lambda m/M`B. `lambda M/m`C. `lambdasqrt(m/M)`D. `lambdasqrt(M/m)` |
| Answer» Correct Answer - 3 | |
| 115. |
Through what potential difference should an electron be accelerated so that its de-Broglie wavelength becomes 0.4Å? |
| Answer» `lambda=12.27/(sqrtV)Å or 0.4=12.27/(sqrtV) or V=(12.27/0.4)^2=941.0V` | |
| 116. |
A particle of mass 4m at rest decays into two particles of masses m and 3m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles 1 and 2 isA. `(1)/(2)`B. `(1)/(4)`C. 2D. 1 |
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Answer» Correct Answer - D (d) : According to law of conservation of linear momentum. The de Broglie wavelength is given by `lamda=(h)/(p)" ":." "(lamda_(1))/(lamda_(2))=1` |
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| 117. |
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. `1.0`D. `sqrt(m_(2))// sqrt(m_(1))` |
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Answer» Correct Answer - C By law of conservation of momentum `0 = m_(1) vec(v_(1)) + m_(2) vec(v_(2)) rArr m_(1) vec (v_(1)) = - m_(2) vec(v_(2))` `- ve` sign indicates that both the particles are moving in opposite direction. Now de - Broglie wavelengths `lambda_(1) = (h)/(m_(1) v_(1))` and `lambda_(2) = (h)/(m_(2) v_(2)), :. (lambda_(1))/(lambda_(2)) = (m_(2) v_(2))/(m_(1) v_(1)) = 1` |
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| 118. |
The de - Broglie wavelength of a neutron at `27 ^(@) C` is `lambda`. What will be its wavelength at `927^(@) C`? |
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Answer» Correct Answer - `lambda_2=lambda/2` `lambda prop 1/sqrtT` |
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| 119. |
The kinetic energy of an electron with de - Broglie wavelength of `0.3 nanometre` isA. `0.168 eV`B. `16.8 eV`C. `1.68 eV`D. `2.5 eV` |
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Answer» Correct Answer - A `lambda = (h)/(sqrt(2 mE)) rArr E = (h^(2))/( 2 m lambda^(2))` `= ((6.6 xx 10^(-34))^(2))/(2 xx 9.1 xx 10^(-31) xx (0.3 xx 10^(-9))^(2))` `= 2.65 xx 10^(-18) J = 16.8 eV` |
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| 120. |
When light of wavelength `300 nm` (nanometre) falls on a photoelectric emitter , however light of `600 nm` wavelength is sufficient for creating photoemission . What is the ratio of the work functions of the two emitters ?A. `1 : 2`B. `2 :1`C. `4 : 1`D. `1 : 4` |
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Answer» Correct Answer - B Work function ` = ( hc)/(lambda_(0))` , where `lambda_(0)` is threshold wavelength . `:. (W_(01))/(W_(02)) = (lambda_(02))/(lambda_(01)) = (2)/(1)` |
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| 121. |
Is photoelectric emission possible at all frequencies? Give reason for your answer? |
| Answer» No, photo -electric emission from a metal surface is possible if the frequency of the incident light is greater than the threshold frequency `(v_0)` for that metal or the energy of the incident photon is greater than the work function of metal, i.e., `vgtv_0`, where `phi_0=hv_0` | |
| 122. |
Out of microwaves, ultraviolet rays and infra-red rays, which radiations will be most effective for emission of electrons from metallic surface? |
| Answer» Out of the given radiations, ultraviolet rays are most energetic and hence effective for emission of electrons from a metallic surface. | |
| 123. |
The minimum energy required for the electron emission from the metal surface can be supplied to the free electrons by which of the following physical processes ?A. Thermionic emissionB. Field emissionC. photoelectric emissionD. All of these |
| Answer» Correct Answer - D | |
| 124. |
Assertion : Kinetic energy of photo electrons emitted by a photosensitive surface depends upon the intensity of incident photon. Reason : The ejection of electrons from metallic surface is possible with frequency of incident photon below the threshold frequency.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 - D According to Einstein equation , `KE = hv - hv_(0)`, i.e., `KE` depends upon the frequency . Photoelectron emitted only if incident frequency is more than threshold frequency. |
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| 125. |
Assertion : Kinetic energy of photo electrons emitted by a photosensitive surface depends upon the intensity of incident photon. Reason : The ejection of electrons from metallic surface is possible with frequency of incident photon below the threshold frequency.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 - D According to Einstein equation `KE = hv - hv_(0) , i.e ., KE` depends upon the frequency . Photoelectron emitted only if incident frequency more than threshold frequency. |
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| 126. |
Assertion : Free electrons inside a metal are free to move out of the metal. Reason : Free electrons inside conductor do not need additional energy to get out of the metal.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. |
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Answer» Correct Answer - D (d) : Free electrons in a metal are free to move inside the metal in a constant. They are not free to move out of the metal. They need additional energy to get out of the metal. |
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| 127. |
Assertion : In photoelctric effect , on increasing the intensity of light , both the number of electrons emitted and kinetic energy of each of them get increased but photoelectric current remains unchanged. Reason : The photoelectric current depends only on wavelength of light .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 - D On increasing the intensity of incident light , the current in photoelectric cell will increase. The energy of the photons `(hv)` will , however not increase with increase in intensity , and hence the kinetic energy of the emitted electrons will not increase. |
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| 128. |
Assertion : Photoelectric current depends on the intensity of incident light. Reason : Number of photoelectrons emitted per second is directly proportional to intensity of incident radiation.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. |
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Answer» Correct Answer - A (a) : Photoelectric current depends on (i) intensity of incident light (ii) The potential difference applied between the two electrodes (iii) the nature of emitted material. |
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| 129. |
If an electron behaves like a wave, what should determine the wavelength and frequency of this wave. |
| Answer» Momentum and Energy. | |
| 130. |
Assertion : The energy `(E )` and momentum `(p)` of a photon are related by `p = E//c`. Reason : The photon behaves like a particle.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 - A Momentum of a photon is given by `y = (h)/(lambda)`. Also the photon is a form of energy packets behave as a particle having energy `E = (hc)/(lambda)`. So `p = (E )/(c )`. |
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| 131. |
Assertion : Photoelectric effect demonstrates the wave nature of light. Reason: The number of photoelectrons is proportional to the frequency of light.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 - D Photoelectric effect demonstrates the particle nature of light. Number of emitted photoelectrons depends upon the intensity of light. |
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| 132. |
The work fuction of caseium is 1.8 eV. Light of 4500Å is incident on it. Calculate (i) the maximum kinetic energy of the emitted photoelectrons (ii) maximum velocity of the emitted photoelectrons (iii) if the intensity of the incident light is doubled, then find the maximum kinetic energy of the emitted photoelectrons. Given, `h=6.63xx10^(-34)Js.m_(e)=9.1xx10^(-31)kg, c=3xx10^(8)ms^(-1)` |
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Answer» Correct Answer - `(i) 1.52xx10^(-19)J (ii) 5.78xx10^(5) ms^(-1) (iii) 1.52xx10^(-19)J` Here, `phi_(0)=1.8eV, lambda=4500 A =4.5xx10^(-7)m` (i) Max K.E. of emitted photoelectrons is `K_(max)=(hc)/lambda-phi_(0)` `=(6.63xx10^(-34)xx3xx10^(8))/(4.5xx10^(-7))-1.8xx1.6xx10^(-19)` `=4.4xx10^(-19)-2.88xx10^(-19)=1.52xx10^(-19)J` (ii) Max. velocity of emitted photoelectron `v_(max)=sqrt((2K_(max))/m)=sqrt((2xx1.52xx10^(-19))/(9.1xx10^(-31)))` `=5.78xx10^(5)ms^(-1)` (iii) Kinetic energy of the emitted photoelectron is independent of the intensity of the incident light. Hence, if intensity of incident light is double the max K.E. of the emitted photoelectron remains unchanged `(=1.52xx10^(-19)J)` |
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| 133. |
The work function of caesium is 2.14 eV. When light of frequency `6xx10^(14)Hz` is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons. (b) stopping potential and (c) maximum speed of the emitted photoelectrons. given , `h=6.63xx10^(-34)Js, 1eV=1.6xx10^(-19)J, c=3xx10^(8)m//s`. |
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Answer» (a) Max. K.E. `=hv-phi_(0)=(6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))-2.14=0.346eV=0.35eV` (b) `eV_(0)=Max. K.E. =0.35 eV or V_(0)=0.35eV` (c) `1/2mv_(max)^(2)=0.346 eV=0.346xx1.6xx10^(-19)J` of `v_(max)=((0.346xx1.6xx10^(-19)xx2)/(9.1xx10^(-31)))^(1//2)=3.488xx10^(5)m//s=348.8 km//s =349kms^(-1)` |
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| 134. |
The work fuction of caseium is 2.14 eV. Find (a) the threshold frequency for caesium and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60V. Given `h=6.63xx10^(-34)Js.` |
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Answer» (a) `v_0=(phi_0)/h=(2.14eV)/(6.63xx10^(-34)Js)` `=(2.14xx1.6xx10^(-19)J)/(6.63xx10^(-34)Js)=5.16xx10^(14)Hz` (b) `eV_0=(hc)/lambda-phi_0 or lambda=(hc)/((eV_0+phi_0))` `:. lambda=((6.63xx10^(-34))xx(3xx10^8m//s))/((exx0.6V+2.14eV))` `= (19.89xx10^(-26)Jm)/(2.74xx1.6xx10^(-19)J)~~454nm` |
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| 135. |
The work function of caesium metal is 2.14 eV . When light of frequency `6xx 10^(14)` Hz is incident on the metal surface , photomission of electrons occurs . What is the maximum kinetic energy of the emitted electrons |
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Answer» Given, `phi_(0)=2.14 eV, v =6xx10^(14)Hz` `KE_("max")=hv-phi_(0)=(6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))-2.14 therefore KE_("max")=0.35 eV` |
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| 136. |
A and B are two metals with threshold frequencies `1.8xx10^(14)Hz and 2.2xx10^(14)Hz`. Two identical photons of energy of 0.825 eV each are incident on them. Then photoelectrons are emitted in take `h=6.6xx10^(-34)J//s`A. B aloneB. A aloneC. neither A nor BD. Both A and B |
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Answer» Correct Answer - B (b) : `phi_(0A)=(hv_(0))/(e)eV=((6.6xx10^(-34))xx(1.8xx10^(14)))/(1.6xx10^(-19))eV` =0.74eV `phi_(0B)=((6.6xx10^(-34))xx(2.2xx10^(14)))/(1.6xx10^(-19))eV=0.91eV` Since the incident energy 0.825 eV is greater than 0.74 eV and less than 0.91 eV, so photoelectrone are emitted from metal A only. |
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| 137. |
Two photons, each of energy 2.5 eV are simultaneously incident on the metal surface. If the work function of the metal is 4.5 eV, then from the surface of metalA. one electron will be emitted with energy 0.5eVB. two electrons will be emitted with energy 0.25 eVC. more than two electrons will be emittedD. not a single electron will be emitted |
| Answer» Correct Answer - d | |
| 138. |
How many photons are received on earth per `cm^-2` per hour if the energy from the sun reaching on the earth is at the rate of `2cal cm^(-2) min^(-1)` and average wavelength of solar light be taken as 5500Å. Use `h=6.6xx10^(-34)J , c=3xx10^8ms^(-1)` and `1cal =4,2J` |
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Answer» Here, `lambda=5500Å =5500xx10^(-10)m` Rate of solar energy reaching earth from the sun `=2calcm^-2min^-1=2xx4.2Jcm^-2min^-1` `=2xx4.2xx60Jcm^-2hour^-1` Energy of photon received on earth `E=(hc)/lambda=((6.6xx10^(-34))xx(3xx10^8))/(5500xx10^(-10))` `=3.6xx10^(-19)J` If n number of photons is reaching the earth per `cm^2`per hour, then total energy will be `=nE=nxx3.6xx10^(-19)J cm^-2h^-1` `:. nxx3.6xx10^(-19)=2xx4.2xx60` or `n=(2xx4.2xx60)/(3.6xx10^(-19))=1.4xx10^(21)` |
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| 139. |
The energy flux of the sunlight reaching the surface of the earth is `1.388xx10^(3) W m^(-2)`. How many photons (nearly) per square meter are incident on the earth per second? Assume that the photonsin the sunlight have an average wavelength of 550 nm. |
| Answer» `n=P/E=P/(hc//lambda)=(Plambda)/(hc)=((1.388xx10^(3))xx550xx10^(-9))/((6.63xx10^(-34))xx(3xx10^(8)))=3.84xx10^(21)"photons"//m^(2)-s` | |
| 140. |
Penetrating power of `X` - rays does not depend onA. WavelengthB. EnergyC. Potential differenceD. Current in the filament |
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Answer» Correct Answer - D When current through the filament increases , number of emitted electrons also increases . Hence intensity of `X` - ray increases but no effect on penetration power. |
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| 141. |
Which of the following is not the property of a cathode raysA. It casts shadowB. It produces heating effectC. It produces fluorescenceD. It does not deflect in electric field |
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Answer» Correct Answer - D Cathode rays are stream of negative charged particle , so they deflect in electric field. |
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| 142. |
When the photons of energy hv fall on a photosensitive metallic surface of work function `hv_(0)`, electrons are emitted are from jthe surface. The most energetic electron coming out of the surfece have kinetic energy equal toA. hvB. `hv_(0)`C. `hv+hv_(0)`D. `hv-hv_(0)` |
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Answer» Correct Answer - D (d) : Maximum kinetic energy of the emitted photoelectrons `=hv-hv_(0)`. |
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| 143. |
Light of wavelength `5000 Å` falls on a sensitive plate with photoelectric work function of `1.9 eV`. The kinetic energy of the photoelectron emitted will beA. `0.58 eV`B. `2.48 eV`C. `1.24 eV`D. `1.16 eV` |
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Answer» Correct Answer - A `E = W_(0) + K_(max) , E = (12375)/(5000) = 2.475 eV` `:. K_(max) = E - W_(0) = 2.475 - 1.9 = 0.57 eV` |
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| 144. |
X-rays of wavelength `lambda` fall on photosenstive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-broglie wavelength of electrons emitted will be `sqrt((hlambda)/(2mc))`. |
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Answer» As work function of a surface can be neglected so `phi_0=0`. Therefore, kinetic energy of emitted photoelectron `1/2mv^2=(hc)/lambda-phi_0=(hc)/lambda or mv=sqrt((2mhc)/lambda)` de-Broglie Wavelength of emitted photoelectron `lambda_1=h/(mv)=h/(sqrt((2mhc)//lambda))=sqrt((hlambda)/(2mc))` proved |
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| 145. |
A source of light is emitting photons. Does all the photons emitted have the same energy? Is the source monochromatic? Explain. |
| Answer» A monochromatic source of light emits photons of single frequency (V). Energy of each photon is hv. The photons emitted as the different atoms of the source of light are excited to different states. Therefore, we can say that a source of light is not a monochromatic source of light. | |
| 146. |
A lamp mainly emits light of wavelength `lambda`. The lamp is rated at P watt and 8% of the energy is emitted as visible light. (i) How many photons of light are emitted by the lamp per second? (ii) How many photons are falling per second on a square whose length of each side is a, held perpendicular to the incident photons at a distance r from the lamp. |
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Answer» (i) Energy of each photon, `E=(hc)/lambda` Light energy emitted per second by lamp `8/100P` No. of photons of light emitted by lamp per second, `N=(8P//100)/(hc//lambda)=(8Plambda)/(100hc)` (ii) No. of phtons falling per second on square `=N/(4pir^2)xxa^2=(8Plambda)/(100hc)xx(a^2)/(4pir^2)=1/50 (Plambdaa^2)/(pihcr^2)` |
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| 147. |
The potential energy of a partical varies as .`U(x) = E_0 ` for ` 0 le x le 1``= 0` for `x gt 1 ` for `0 le x le 1` de- Broglie wavelength is `lambda_1` and for `xgt1` the de-Broglie wavelength is `lambda_2`. Total energy of the partical is `2E_0`. find `(lambda_1)/(lambda_2).`A. `2`B. `1`C. `sqrt(2)`D. `(1)/(sqrt(2))` |
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Answer» Correct Answer - C `K.E. = 2 E_(0) - E_(0) = E_(0) (for 0le x le 1)` `rArr lambda_(1) = (h)/(sqrt( 2 m E_(0))` `K.E. = 2 E_(0) ( for x gt 1) rArr lambda_(2) = (h)/(sqrt( 4 m E_(0))) rArr (lambda_(1))/(lambda_(2)) = sqrt(2)` |
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| 148. |
The continuous `X` - rays spectrum produced by an `X` - ray machine at constant voltage hasA. A maximum wavelengthB. A minimum wavelengthC. A single wavelengthD. A minimum frequency |
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Answer» Correct Answer - B Continuous spectrum of `X - rays` consists of radiations of all possible wavelength range having a definite short wavelength limit. |
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| 149. |
Molybdenum is used as a target element for production of `X` - rays because it isA. A heavy element and can easily absorb high velocity electronsB. A heavy element with a high melting pointC. An element having high thermal conductivityD. Heavy and can easily deflect electrons |
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Answer» Correct Answer - B In `X - ray` tube , target must be heavy element with high melting point. |
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| 150. |
If the kinetic energy of the particle is increased to `16` times its previous value , the percentage change in the de - Broglie wavelength of the particle isA. `25`B. `75`C. `60`D. `50` |
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Answer» Correct Answer - B As we know `lambda = (h)/(p) = (h)/(sqrt(2 mK))` `(because P = sqrt( 2mKE))` or `(lambda_(1))/(lambda_(2)) = sqrt((K_(2))/(K_(1))) = sqrt((16 K)/( K)) = (4)/(1)` Therefore the percentage change in de - Broglie wavelength ` = ( 1 - 4)/(4) xx 100 = - 75%` |
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