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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.
| 151. |
Find the energy that should be added to an electron of energy 2eV to reduce its de-Broglie wavelength from 1nm to 0.5nm. |
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Answer» Correct Answer - `6 eV` As `lambda=h/(sqrt(2mK)) so K=(h^(2))/(2mlambda^(2)) or K prop 1/(lambda^(2))` `:. (K_(1))/(K_(2))=[(lambda_(2))/(lambda_(1))]^(2)=[0.5/1]^(2)=1/4` or `K_(2)=4K_(1)=4xx2eV=8eV` Increases in energy `=K_(2)-K_(1)=8-2=6eV` |
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| 152. |
(a) The work function for the surface of aluminium is 4.2eV. Find potential difference will be required to stop the emission of maximum kinetic energy photoelectrons emitted by light of 1800Å wavelenght? (b) Determine the wavelength of that incident light for which stopping potential will be zero? `(h=6.6xx10^(-34))` |
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Answer» (a) Here, `phi_0=4.2eV=4.2xx1.6xx10^(-19)J`, `lambda=1800Å=1800xx10^(-10)m=18xx10^-8m`, `V_0=?` Maximum KE of the emitted photoelectron `K_(max)=(hc)/lambda-phi_0=((6.6xx10^(-34))xx(3xx10^8))/(18xx10^-8)` `-4.2xx1.6xx10^(-19)` `=11xx10^(-19)-6.72xx110^(-19)=4.28xx10^(-19)` Stopping potential, `V_0=(K_(max))/l=(4.28xx10^(-19))/(1.6xx10^(-19))=2.675V` |
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| 153. |
(a) The work function for the surface of aluminium is 4.2eV. How much potential difference will be required to stop the emission of maximum energy electrons emitted by light of 2000Å wavelenght? (b) What will be the wavelength of that incident light for which stopping potential will be zero ? `h=6.6xx10^(-34) Js, c=3xx10^8ms^-1`. |
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Answer» Here, `phi_0=4.2eV` `=4.2xx1.6xx10^(-19)J,` `V=?, lambda=2000Å=2000xx10^(-10)m` Max. K.E. of the emitted photoelectron, `K_(max)=(hc)/lambda-phi_0` `(6.6xx10^(-34)xx3xx10^8)/(2000xx10^(-10))-4.2xx1.6xx10^(-19)J` `=3.18xx10^(-19)J` stopping potential, `V_0=(K_(max))/e=(3.18xx10^(-19))/(1.6xx10^(-19))` `=1.9875V` (b) For threshold wavelength `lambda_0`, the stopping potential is zero. `:. lambda_0=(hc)/(phi_0)=(6.6xx10^(-34)xx3xx10^8)/(4.2xx1.6xx10^(-19))` `=2.946xx10^-7m` `=2946Å` |
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| 154. |
Given that a photon of light of wavelength 10,000Å has an energy equl to 1.23eV. When light of wavelength 5000Å and intensity `I_(0)` falls on a photoelectric cell and the saturation current is `0.40xx10^(-6)` ampere and the stopping potential is 1.36 volt, then (i)what is the work function? (ii) If intensity of light is made `4I_(0)`, what should be the saturation current and stopping potential? |
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Answer» Correct Answer - (i) 1.1 eV (ii) `1.6xx10^(-6) A`, unchanged (i) `(E_(2))/(E_(1))=(hc//lambda_(2))/(hc//lambda_(1))=(lambda_(1))/(lambda_(2))=10000/5000=2` or `E_(2)=2E_(1)=2xx1.23=2.46 eV` `:.` Energy of incident photon, E=2.46 eV, Given, stopping potential =1.36 V, therefore, Kinetic energy of photoelectrons, `K=1.36 eV`. So, work function is `phi_(0)=E-K=2.46-1.36=1.1 eV` (ii) When the intensity of the light is made four times, the stopping potential remains unchanged (i.e., 1.36 eV) but saaturation current will become fout times `=4xx0.40xx10^(-6)=1.6xx10^(-6)A` |
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| 155. |
Light of wavelength 5000Å falls on a metal surface of work function 1.9eV. Find (i) the energy of photons in eV (ii) the kinetic energy of photoelectrons and (iii) the stopping potential. Use `h=6.63xx10^(-34)Js`, `c=3xx10^8ms^-1 , e=1.6xx10^-9C.` |
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Answer» Here, `lambda=5000Å=5xx10^-7m`, `phi_0=1.9eV` (i) Energy of photon in eV, `E=(hc)/lambda=((6.63xx10^(-34))xx(3xx10^8))/((5xx10^-7)xx(1.6xx10^(-19)))eV` `=2.4825eV` (ii) Kinetic energy of photoelectron `=(hc)/lambda-phi_0=2.4825-1.9` `=0.5825eV` (iii) If `v_0` is the stopping potential, then `eV_0=K.E.` of emitted photoelectron =0.5825eV or `V_0=(0.5825eV)/e=0.5825V` |
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| 156. |
Name the device that converts changes in intesity illumination into changes in electric current. Give three applications of this device. |
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Answer» A photoelectric cell converts change in intensity illumination into changes in electric current. The application of photoelectric cells are (i) in burglar alaram (ii) in fire alaram (iii) in the reproduction of sound from films in cinema hall. |
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| 157. |
The work functions for sodium and copper are `2 eV` and `4 eV` . Which of them is suitable for a photocell with `4000 Å` light ?A. CopperB. SodiumC. BothD. None of these |
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Answer» Correct Answer - B Threshold wavelength for `Na` , `lambda_(Na) gt (12375)/(2) = 6187.5 Å` Also `lambda_( cu) = ( 12375)/( 4) = 3093.75` Since `lambda_(Na) gt 4000 Å` , So `Na` is suitable. |
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| 158. |
Two identical photocathodes receive light of frequency `f_(1) andf_(2)` if the velocites of the photo electrons (of mass `m`) coming out are repectively `v_(1) and v_(2)` thenA. `v_(1) - v_(2) = [ ( 2h)/(m) (f_(1) - f_(2))]^(1//2)`B. `v_(1)^(2) - v_(2)^(2) = ( 2h)/(m) (f_(1) - f_(2))`C. `v_(1) + v_(2) = [ ( 2h)/(m) (f_(1) + f_(2))]^(1//2)`D. `v_(1)^(2) + v_(2)^(2) = ( 2h)/(m) (f_(1) + f_(2))` |
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Answer» Correct Answer - B Using Einstein photoelectric equation `E = W_(0) + K_(max)` `hf_(1) = W_(0) + (1)/(2) mv_(1)^(2)` ….(i) `hf_(1) = W_(0) + (1)/(2) mv_(2)^(2)` …..(ii) `rArr h(f_(1) - f_(2)) = (1)/(2) mv(v_(1)^(2) - v_(2)^(2))` `rArr (v_(1)^(2) - v_(2)^(2)) = (2h)/(m) (f_(1) - f_(2))` |
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| 159. |
(i) In the explanations of photoelectric effect, we assume one photon of frequency `nu` collides with an electron and transfer its energy. This leads to the equation for the maximum energy `E_(max)` of the emitted electron as `E_(max)=hnu-phi_(0)` Where `phi_(0)` is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron? (ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential? |
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Answer» Here it is given that, an electron absorbs 2 photons each of frequency v then v = 2v where, v is the frequency of emitted electron. Given, `" "E_(max)=hv-phi_(0)` Now, maximum energy for emitted electrons is `" "E_(max)=h(2v)-phi_(0)-2hv-phi_(0)` (ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible. |
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| 160. |
Consider figure. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of `theta` that A. will be larger than the earlier valueB. will be the same as the earlier valueC. will be less than the earlier valueD. will depend on the target |
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Answer» In Davisson-Germer experiment the de-Broglie wavelength associated with electron is `" "lamda=(12.27)/(sqrtV)Å` where V is the applied valtage. If there is a maxima of the diffracted electrons at an angle `theta,` then `" "2dsintheta=lamda` From Eq. (i). we note that if V is inversely proportional to the wavelength `lamda.` i.e., V will increase with the decrease in the `lamda.` From eq. (ii), we note that wavelength `lamda` is directly proportional to `sintheta` and hence `theta.` So, with the decrease in `lamda,theta` will also decreases. Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum at a value of `theta` that will be less than the earlier value. |
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| 161. |
Consider the fundamental physics for class XII. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of `theta` thatA. will be larger then that earlier valueB. will be the same as the earlier valueC. will be less than the earlier valueD. will depend on the target |
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Answer» Correct Answer - c In Davisson and Germer Experiment, if V is the voltage applied to anode, then de-Broglie wavelength associated with electron is `lambda=12.27/(sqrt(V))Å.......(i)` If there is a maxima of the diffracted electrons at an angle `theta`, then `2d sin theta=lambda.........(ii)` From (ii), we note that as `lambda` decreases, the value of `sin theta` decreases, i.e., `theta` decreases. |
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| 162. |
Consider a beam of electrons (each electron with energy `E_(0))` incident on a metal surface kept in an evacuated chamber. ThenA. no electrons will be emitted only photons can emit electronsB. electrons can be emitted but all with an energy, `E_(o)`C. electrons can be emitted with any energy, with a maximum of `E_(o)-phi` (`phi` is the work function)D. electrons can be emitted with any energy, with a maximum of `E_(o)` |
| Answer» When a bean of electrons of energy `E_(o)` is incident on a metal surface kept in an evaacuated chamber electrons can be emitted with maximum energy `E_(o)` (due to elastic collidion) and with any energy less then `E_(o),` when part of incident energy of electron is used in liberating the electrons from the surface of metal. | |
| 163. |
Consider a beam of electrons (each electron with energy `E_(0))` incident on a metal surface kept in an evacuated chamber. ThenA. no electrons will be emitted as only photons can emit electrons.B. electron can be emitted but all with an energy `E_(0)`.C. electrons can be emitted with any energy, with a maximum of `E_(0)-phi(phi"is the work function")`.D. electrons can be emitted with energy, with a maximum of `E_(0)`. |
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Answer» Correct Answer - D (d) : when a beam of electrons of energy `E_(0)` is incident on a metal surface kept in evacuated chamber, elecrtrons can be emitted with maximum energy `E_(0)` (due to elastic collision) and with any energy less than `E_(0)`, when part of incident ebergy of electron is used in liberating electrons from the surface of metal. |
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| 164. |
A neutron is an unchanged particle of mass `1.67xx10^(-27) kg`. Calculate the de-Broglie wavelength of the neutron moving with a velocity, such that K.E. is 0.04 eV, `h=6.62xx10^(-34)Js`. |
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Answer» Correct Answer - `1.43Å` K.E. of neutron, `K=1/2mv^(2) or mv=sqrt(2mK)` `lambda=h/(mv)=h/(sqrt(2mK))` |
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| 165. |
An electron is accelerated through a potential difference of 64volts. What is the de-broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? |
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Answer» Correct Answer - `1.53Å`, x-rays` `lambda=(12.27Å)/(sqrt(64))=1.53Å` This wavelength belong to x-rays. |
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| 166. |
If the wavelength of light incident on a photoelectric cell be reduced from `4000Å` to `3600Å`, then what will be the change in the cut off potential? `(h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C)` |
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Answer» Let `phi_0` be the work function of the surface of the photoelectric cell to be exposed by light. Maximum KE of emitted photoelectron will be `K_(max)=(hc)/lambda-phi_0` If the `V_0` is the cut off potential, then `eV_0=K_(max)=(hc)/lambda-phi_0` or `V_0=(hc)/(e lambda)-(phi_0)/e` When the incident light of wavelength `lambda_1` be changed to wavelength `lambda_(2)`, then the change (increase) in cut off potential will be `DeltaV_0=((hc)/(elambda_2)-(phi_0)/e)-((hc)/(elambda_1)-(phi_0)/e)` `=(hc)/e (1/(lambda_1)-1/(lambda_1))` `=((6.6xx10^(-34))xx(3xx10^8))/((1.6xx10^(-19)))[(10^(10))/3600-(10^(10))/4000]` `=0.34V` |
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| 167. |
The graph of Fig. shows the variation of photoelectric current (I) versus applied voltage (V) for the two different photosensitive materials for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. |
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Answer» The similar materials will have same stopping potential and the different materials will have different stopping potential whatever may be the intensity of incident radiation. For the same intensity of two different incident radiations on two different materials, there will be same value of photoelectric current. So the pair of the curves (1 and 3) or (2and 4) correspond to different materials but having same intensity of incident radiations. |
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| 168. |
Calculate the energy of a photon of blue light, `lambda`=450 nm in air (or vacuum). |
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Answer» Strategy : The photon has energy E=hf , where `f=c//lambda` Since `f=c//lambda`, we have `E=hf="hc"/lambda=((6.63xx10^(-34)J.s)(3.0xx10^8 m//s))/((4.5xx10^(-7)m))=4.4xx10^(-19) `J, or `((4.4xx10^(-19) J))/((1.60xx10^(-19) J//eV))` =2.8 eV 1 eV=`1.60xx10^(-19)` J |
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| 169. |
A photon of energy `8 eV` is incident on a metal surface of threshold frequency `1.6 xx 10^(15) Hz`, then the maximum kinetic energy of photoelectrons emitted is `( h = 6.6 xx 10^(-34) Js)`A. `4.8 eV`B. `2.4 eV`C. `1.4 eV`D. `0.8 eV` |
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Answer» Correct Answer - C Work function `W_(0) = hv_(0) = 6.6 xx 10^(-34) xx 1.6 xx 10^(15)` `= 1.056 xx 10^(-18) J = 6.6 eV` From `E = W_(0) + K_(max) rArr K_(max) = E - W_(0) = 1.4 eV` |
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| 170. |
The threshold wavelength for photoelectric emission for a material is `5200 A^(@)`. Will the photoelectrons be emitted when this material is illuminated with monochromatic radiation from 1 watt ultra violet lamp? |
| Answer» Yes, because wavelength of ultra violet light is less than the threshold wavelength 5200Å. | |
| 171. |
Light of frequency `4 v_(0)` is incident on the metal of the threshold frequency `v_(0)`. The maximum kinetic energy of the emitted photoelectrons isA. `3 hv_(0)`B. `2 hv_(0)`C. `(3)/(2) hv_(0)`D. `(1)/(2) hv_(0)` |
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Answer» Correct Answer - A `E = hv_(0) + K_(max) rArr K_(max) = 3 hv_(0)`. |
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| 172. |
Find the threshold wavelength of light that would produce photoelectrons for a silver surface . The work function for silver is 4.8 eV |
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Answer» Energy conservation for the photoelectric effect gives `eV_0=hv-phi` At threshold, the photoelectrons have zero kinetic energy, Solving for the wavelength yields `lambda="hc"/phi`=259 nm |
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| 173. |
The essential distinction between `X` - rays and `gamma` - rays is thatA. `gamma` - rays have smaller wavelength than `X` - raysB. `gamma` - rays emanate from nucleus while `X` - rays emanate from outer part of the atomC. `gamma` - rays have greater ionizing power than `X` - raysD. `gamma` - rays are more penetrating than `X` - rays |
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Answer» Correct Answer - B The wavelength of the `gamma` - rays is shorter . However the main distinguishing feature is the nature of emission. |
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| 174. |
By which way, the X-rays, and `gamma`-rays can be distinguised?A. Their velocityB. Their ionising powerC. Their intensityD. Method of production |
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Answer» Correct Answer - D The production of `X - rays` is an atomic property whereas the production of `gamma - rays ` is a nuclear property . |
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| 175. |
Assertion : Mass of moving photon varies inversely as the wavelength . Reason : Energy of the particle `= Mass xx (Speed of light)^(2)`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 Mass of moving photon `m = (hv)/(c^(2))` and `E = mc^(2)`. |
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| 176. |
An electron of mass `m` with an initial velocity `vec(v) = v_(0) hat`(i) `(v_(0) gt 0)` enters an electric field `vec(E ) =- E_(0) hat (i)` `(E_(0) = constant gt 0)` at `t = 0` . If `lambda_(0)` is its de - Broglie wavelength initially, then its de - Broglie wavelength at time `t` isA. `lamda_(0)`B. `lamda_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))`C. `lamda_(0)/(sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2))))`D. `lamda_(0)/((1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2))))` |
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Answer» Correct Answer - C (c ) : Initial de Broglie wavelength of electron, `lamda_(0)=(h)/(mv_(0))` Force on electron in electric field, `vec(F)=-evec(E)=-eE_(0)hat(j)` Acceleration of electron, `vec(a)=(vec(F))/(m)=-(eE_(0))/(m)hat(j)` It is acting along negative along x-axis `vec(v_(x0))=v_(0)hat(i)`. Initial velocity of electron along y-axis `vec(v_(y0))=v_(0)hat(i)` `( :.` there is no acceleration of electron along x-axis) Velocity of electron after time t y- axis, `vec(v_(y))=0+(-(eE_(0))/(m)hat(j))t=-(eE_(0))/(m)that(j)` Magnitude of velocity of elecron after time t is `|vec(v)|=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(v_(0)^(2)+((-eE_(0))/(m)t)^(2))=v_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))` de Broglie wavelength associated with electron at time t is `lamda=(h)/(mv)=(h)/(mv_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/((m^(2)v_(0)^(2)))))=(lamda_(0))/sqrt(1+(e^(2)E_(0)^(2)t^(2))/((m^(2)v_(0)^(2))))` |
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| 177. |
An electron of mass `m` with an initial velocity `vec(v) = v_(0) hat`(i) `(v_(0) gt 0)` enters an electric field `vec(E ) =- E_(0) hat (i)` `(E_(0) = constant gt 0)` at `t = 0` . If `lambda_(0)` is its de - Broglie wavelength initially, then its de - Broglie wavelength at time `t` isA. `(lambda_(0))/((1+(eE_(0))/(m)(t)/(v_(0))))`B. `lambda_(0)(1+(eE_(0)t)/(mv_(0)))`C. `lambda_(0)`D. `lambda_(0)t` |
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Answer» Correct Answer - a Initial de-Broglie wavelength of electron, `lambda_(0)=h/(mv_(0)).......(i)` Force on electron in electric field, `vecF=-evecE=-e[-E_(0)hati]=eE_(0)hati` Acceleration of electron, `veca=(vecF)/m=(eE_(0)hati)/m` velocity of electron after time t, `vecv=v_(0)hati+((eE_(0)hati)/m)t=(v_(0)+(eE_(0))/mt)hati=v_(0)(1+(eE_(0))/(mv_(0))t)hati` De-Broglie wavelength associated with electron at time t is `lambda=h/(mv)=h/(m[v_(0)(1+(eE_(0))/(mv_(0))t)])=(lambda_(0))/([1+(eE_(0))/(mv_(0))t])` [from (i)] |
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| 178. |
An electron of mass `m` with an initial velocity `vec(v) = v_(0) hat`(i) `(v_(0) gt 0)` enters an electric field `vec(E ) = v_(0) hat (i) `(E_(0) = constant gt 0)` at `t = 0` . If `lambda_(0)` is its de - Broglie wavelength initially, then its de - Broglie wavelength at time `t` isA. `lambda_(0)`B. `(lambda_(0))/(( 1 + (e E_(0))/(mv_(0)) t)`C. `lambda_(0) t`D. `lambda_(0) (1 + ( e E_(0))/(mv_(0)) t)` |
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Answer» Correct Answer - B `vec(v) = v_(0) hat (i)` `vec ( E ) = - E_(0) hat (i)` `vec (a) = vec(F)/(m) = (q vec( E))/(m) = ((-e) (-E_(0)) hat (i))/(m)` ` rArr vec (a) = ( e E_(0))/(m) hat (i)` `v = u + at` `v = v_(0) + (e E_(0))/(m) t` `lambda = (h)/(mv) = (h)/(m [ v_(0) + (e E_(0) t)/(m)])` `= (h)/(mv_(0) [ 1 + ( eE_(0) t)/(mv_(0))]) = (lambda_(0))/(1 + (e E_(0) t)/(mv_(0))` |
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| 179. |
An electron (mass m) with an initial velocity `v=v_(0)hat(i)(v_(0)gt0)` is in an electric field `E=-E_(0)hat(l)(E_(0)="constant"gt0)`. Its de-Broglie wavelength at time t is given byA. `(lamda_(0))/((1+(eE_(o))/(m)(t)/(v_(0))))`B. `lamda_(0)(1+(eE_(0)t)/(mv_(0)))`C. `lamda_(0)`D. `lamda_(0)t` |
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Answer» Initial de-Broglie wavelength of electron, `" "lamda_(0)=(h)/(mv_(0))` Force on electron in electric field, `" "F=-eE=-e[-E_(0)hati]=eE_(0)hati` Acceleration of electron `" "a=F/m=(eE_(0)hati)/(m)` Vfelocity of electron after time t, `" "v=v_(0)hati+((eE_(0)hati)/(m))t=(v_(0)+(eE_(0))/(m)t)hati` `" "v_(0)(1+(eE_(0))/(mv_(0)))hati` de-Broglie wavelength associated with electron at time t is `" "lamda =(h)/(mv)` `implies" "=(h)/(m[v_(0)(1+(eE_(0))/(mv_(0))t)])=(lamda_(0))/([1+(eE_(0))/(mv_(0))])" "[thereforelamda_(0)=(h)/(mv_(0))]` |
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| 180. |
Assertion : If different gases are filled turn by turn at the same pressure in the discharge tube the discharge in them takes place at the same potential. Reason : The discharge depends only on the pressure of discharge tube and not on the ionisation potential of gas.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 The discharge depends on both pressure of discharge tube and ionisation potential of gas . Since the ionisation potential of different gases are different, hence the discharge in different gases takes place at different potential . |
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| 181. |
A point source of light of constant power P is placed at centre of curvature of a concave mirror. If angle substented by aperature at centre of curvature is `120^@`, calculate force an surface of mirror. |
| Answer» Correct Answer - `"3P"/"8C"` | |
| 182. |
Two particles A and B of de-broglie wavelength `lambda_(1) and lambda_(2)` combine to from a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one dimensional). |
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Answer» `|vecp_(C)|=|vecp_(A)|+|vecp_(B)| or h/(lambda_(C))= h/(lambda_(A))+ h/(lambda_(B)).......(i)` Following cases may arise Case (i) `p_(A)gt0, p_(B)gt0`, i.e., both `p_(A) and p_(B)` are positive. From (i), `h/(lambda_(C))=(h(lambda_(A)+lambda_(B)))/(lambda_(A)lambda_(B)) or (lambda_(C))=((lambda_(A)lambda_(B)))/(lambda_(A)+lambda_(B))` Case (ii) `p_(A)lt0, p_(B)lt0`, i.e., both `p_(A) and p_(B)` are negative. From (i), `h/(lambda_(C))=-h/(lambda_(A))-h/(lambda_(B))=- (h(lambda_(A)+lambda_(B)))/(lambda_(A)lambda_(B)) or (lambda_(C))=-((lambda_(A)lambda_(B)))/(lambda_(A)+lambda_(B))` Case (iii) `p_(A)gt0, p_(B)lt0`, i.e., both `p_(A)is +ve and p_(B)` -ve. From (i), `h/(lambda_(C))=h/(lambda_(A))-h/(lambda_(B))= ((lambda_(B)-lambda_(A))h)/(lambda_(A)lambda_(B)) or (lambda_(C))=((lambda_(A)lambda_(B)))/(lambda_(B)-lambda_(A))` Case (iv) `p_(A)lt0, p_(B)gt0`, i.e., both `p_(A)is -ve and p_(B)` +ve. From (i), `h/(lambda_(C))=-h/(lambda_(A))+h/(lambda_(B))= ((lambda_(A)-lambda_(A))h)/(lambda_(A)lambda_(B)) or (lambda_(C))=((lambda_(A)lambda_(B)))/(lambda_(A)-lambda_(B))` |
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| 183. |
What is the Be Broglie wavelength of an electron with kinetic energy of 120 eV. |
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Answer» Given, `KE = 120 eV, m = 9.1xx10^(-3)kg, e = 1.6xx10^(-19)c` `lambda = (12.27)/(sqrt(V)) Å = (12.27)/(sqrt(120)) Å = 0.112xx10^(-9) m therefore lambda = 0.112 nm`. |
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| 184. |
The wavelength `lambda_(e )` of an photon of same energy `E` are related byA. `lambda_(p) prop lambda_(e )^(2)`B. `lambda_(p) prop lambda_( e )`C. `lambda_(p) prop sqrt(lambda_(e ))`D. `lambda_(p) prop (1)/(sqrt(lambda_(e)))` |
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Answer» Correct Answer - A `lambda_(e) = (h)/(sqrt( 2 mE)) lambda_(e) = (hc)/( E)` `lambda_(e)^(2) = (h^(2))/( 2m E)` `lambda_(e)^(2) = (h^(2))/( 2 m (hc)/(lambda_(p))) rArr lambda_( e)^(2) prop lambda_(p)` |
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| 185. |
The de - Broglie wavelength associated with the particle of mass `m` moving with velocity `v` isA. `h//mv`B. `mv//h`C. `h//2v`D. `m//hv` |
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Answer» Correct Answer - A `lambda = (h)/(p) = (h)/(mv)` |
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| 186. |
If particles are moving with same velocity , then maximum de - Broglie wavelength will be forA. NeutronB. ProtonC. `beta`- particleD. `alpha` - particle |
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Answer» Correct Answer - C `lambda = (h)/(mv) rArr prop (1)/(m)` |
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| 187. |
Particle nature and wave nature of electromagnetic waves and electrons can be shown byA. Electron has small mass , deflected by the metal sheetB. X - ray is diffracted , reflected by thick metal sheetC. Light is refracted and diffractedD. Photoelectricity and electron microscopy |
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Answer» Correct Answer - D In photoelectric effect particle nature of electron is shown . While in electron microscope , beam of electron is considered as electron wave. |
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| 188. |
If an electron and a photon propagate in the form of waves having the same wavelength , it implies that they have the sameA. EnergyB. MomentumC. VelocityD. Angular momentum |
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Answer» Correct Answer - B If an electron and a photon propagates in the form of waves having the same wavelength , it implies that they have same momentum . This is according to de - Broglie equation , `p prop (1)/(lambda)` |
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| 189. |
Given in fig. is the graph between frequency v of the incident light and maximum kinetiec energy `(E_(k))` of the emitted photoelectrons. Find the values of (i) threshold frequency and (ii) work function form the graph. |
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Answer» Correct Answer - `(i) 10^(15)Hz (ii) 4eV` (i) Threshold frequency , `v_(0)=10xx10^(14)Hz =10^(15)Hz` (ii) At, `v=0, E_(k)=hxx0-phi_(0)=-phi_(0)` or `phi_(0)=-E_(k)=-(-4eV)=4eV` |
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| 190. |
Many public building have doors that open automatically as people approach. What is the basic concept involved in it? |
| Answer» The automatic opening of the doors in a building is often operated by "electric eyes". On the door of a public bulding, a light beam from a strong source of light shines across the door opening on to a photo voltaic cell. It causes the ejection of electrons, resulting in photoelectric current in the circuit which keeps the door closed. When some person approaches the door for entering into the building, the light beams is prevented from falling on photovotaic cell. Due to it, the current stops in the cell circuit. A mechanism is triggered that opens the door automatically. | |
| 191. |
Why photographer use red light and not blue light while developing the exposed films? |
| Answer» The energy of red light photon is lower than that of blue light photon, i.e., the energy of red light is less than that of the blue light. Therefore, the red light can not cause a chemical change in the exposed film whereas blue light can. That is why the red light is used by photographers while developing the exposed films. | |
| 192. |
"Know your face beauty through complexion meter" was one of the stall on a science exhibition. A student interested to know his/her face beauty was made to stand on a platform and light from a lamp was made to fall on his/her face. The reading of complexion meter indicated the face beauty of the student which might be very fair, fair, semifair, semidark and dark, etc. Read abve passage and answer the following questions:(i) What is the basic concept used in the working of complexion meter? (ii) How is the face beauty recorded by face complexion meter? (iii) What basic values do you learn from the above study? |
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Answer» (i) The basic concept used in the working of complexion meter is photoelectric effects. (ii) The reading of complexion meter depends on the current generated from the reflected from the face. Fairer the coulor more is the light reflected and vice versa. (iii) Complexion meters judge your face beauty only. The inner beauty of a person is judged by the virtues he or she possesess. In may opinion, inner beauty is much more valuable than the face beauty. |
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| 193. |
The energy of a photon of characteristic X-ray from a Coolidge tube comes fromA. The kinetic energy of the striking electronB. The kinetic energy of the free electrons of the targetC. The kinetic energy of the ions of the targetD. An electronic transition of the target atom |
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Answer» Correct Answer - D The energy of `X` - rays photon obtained from a Coolidge tube by an electronic transition from `L` orbit in `K` orbit. |
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| 194. |
The de Broglie wavelength of an electron in a metal at `27^(@)C` is `("Given"m_(e)=9.1xx10^(-31)kg,k_(B)=1.38xx10^(-23)JK^(-1))`A. `6.2xx10^(-9)m`B. `6.2xx10^(-10)m`C. `6.2xx10^(-8)m`D. `6.2xx10^(-7)m` |
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Answer» Correct Answer - A (a) : Here, T=27+273=300K For an electron in a metal, momentum `p=sqrt(3mk_(B)T)` de Broglie wavelength of an electron is `lamda=(h)/(p)=sqrt(3mk_(B)T)` `=(h)/sqrt(3xx(9.1xx10^(-31))xx(1.38xx10^(-23))xx300)` `=6.2xx10^(-9)m` |
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| 195. |
A `alpha` -parhticle moves in a circular path of radius `0.83 cm` in the presence of a magnetic field of `0.25 Wb//m^(2)`. The de-Broglie wavelength assocaiated with the particle will beA. `1Ã…`B. `0.1Ã…`C. `10Ã…`D. `0.01Ã…` |
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Answer» Correct Answer - D (d) : Radius of the circular path of a charged particle in a magnetic field is given by `R=(mv)/(Bq)ormv=RBq` Here, `R=0.83cm=0.83xx10^(-2)m` `B=0.25"Wb m"^(-2)` `q=2e=2xx1.6xx10^(-19)C` `:.mv=(0.83xx10^(-2))(0.25)(2xx1.6xx10^(-19))` de Broglie wavelength, `lamda=(h)/(mv)=(6.6xx10^(-34))/(0.83xx10^(-2)xx0.25xx2xx1.6xx10^(-19))=0.01Ã…` |
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| 196. |
`K_alpha` wavelength emitted by an atom of atomic number Z=11 is `lambda`. Find the atomic number for an atom that emits `K_alpha` radiation with wavelength `4lambda`. (a) Z=6 (b) Z=4 (c ) Z=11 (d) Z=44.A. `Z = 6`B. `Z = 4`C. `Z = 11`D. `Z = 44` |
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Answer» Correct Answer - A `sqrt(f_(1)) = sqrt((v)/(lambda_(1))) = a(11 - 1)` and `sqrt(f_(2)) = sqrt((v)/(lambda_(2))) = a(Z - 1)` "By dividing" , `sqrt((lambda_(1))/(lambda_(2))) = (10)/(Z - 1) rArr sqrt((4)/(1)) = (10)/(Z - 1) rArr Z = 6` |
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| 197. |
A modern 200 W sodium street lamp emits yellow light of wavelength 0.6 `mum`. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second isA. `3 xx 10^(19)`B. `1.5 xx 10^(20)`C. `6 xx 10^(18)`D. `62 xx 10^(20)` |
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Answer» Correct Answer - B Energy of a photon ` = (hc )/(lambda)` The number of photons emitted per second ` = (N)/(t)` Hence power emitted `p = (N)/(t) ((hc)/(lambda))` It is given that sodium lamp has `25%` efficient in converting electricial energy to light energy emitted `P = 200 xx (25)/(100) = 50 W = 50 J//sec` No. of photons `(N)/(t) = (P lambda)/(lambda)` ` = (50 xx 0.6 xx 10^(-6))/(6.6 xx 10^(-34) xx 3 xx 10^(8)) = 1.5 xx 10^(20)` |
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| 198. |
The photo-electrons emitted from a surface of sodium metal are such thatA. All are of the same frequencyB. Have the same kinetic energyC. Have the same de Broglie wavelengthD. Have their speeds varying from zero to a certain maximum |
| Answer» Correct Answer - 4 | |
| 199. |
A photo cell is receiving light from a source placed at a distance of `1 m`. If the same source is to be placed at a distance of `2 m` , then the ejected electronA. Moves with one - fourth energy as that of the initial energyB. Moves with one - fourth of momentum as that of the initial momentumC. will be half in numberD. will be one - fourth in number |
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Answer» Correct Answer - D Number of ejected electrons `prop (Intensity) prop (1)/((Distance)^(2))` Therefore an increament of distance two times will reduce the number of ejected electrons to `(1)/(4) th` of the previous one. |
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| 200. |
A photo cell is receiving light from a source placed at a distance of `1 m`. If the same source is to be placed at a distance of `2 m` , then the ejected electronA. Moves with one-fourth energy as that of the initial energyB. Moves with one-fourth of momentum as that of the initial momentumC. Will be half in numberD. Will be one-fourth in number |
| Answer» Correct Answer - 4 | |