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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. |
`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. |
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Answer» Correct Answer - A `(1)/(lambda) prop (Z-1)^2` `:. (lambda_1)/(lambda_2) = ((Z_2- 1)/(Z_1 - 1))^2 or (1)/(4) = ((Z_2 -1)/(11-1))^2` Solving this , we get Z_2 = 6 `:.` Correct answer is (a). |
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| 202. |
Match the following table. |
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Answer» Correct Answer - A::B::C::D |
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| 203. |
Regarding trasnsition of electrons match the following table |
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Answer» Correct Answer - A::B::C::D |
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| 204. |
`sqrt(v)` versus Z graph for dcharacteristic X-rays is as shown in figure. Mathc the following |
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Answer» Correct Answer - A::B::C::D |
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| 205. |
The decay constant of a radioactive sample is `lamda.` Its half -life is `T_(1//2)` and mean life is T.A. `T_(1//2)=(1)/(lamda),T=(In2)/(lamda)`B. `T_(1//2)=(In2)/(lamda),T=(1)/(lamda)`C. `T_(1//2)=lamda"in" 2, T=(1)/(lamda)`D. `T_(1//2)=(lamda)/(In2),T=(In2)/(lamda)` |
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Answer» Correct Answer - b |
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| 206. |
In a photoelectric experiment, with light of wavelength `lambda`, the fastest election has speed v. If the exciting wavelength is changed to `(3lambda)/4`, the speed of the fastest emitted electron will becomeA. `= v((4)/(3))^(1/2)`B. `= v((3)/(4))^(1/2)`C. `gt v((4)/(3))^(1/2)`D. `lt v((4)/(3))^(1/2)` |
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Answer» Correct Answer - C `hv_(0)^(2) - hv_(0) = (1)/(2) m v^(2)` `:. (4)/(3) hv_(0) - hv_(0) = (1)/(2) m v^(-2)` `:. (v^(2))/(v^(2) = ((4)/(3) v - v_(0)))/(v - v_(0)) :. V ^(1) = v sqer(((4)/(3) v - v_(0)))/(v - v_(0)))` `:. V^(1) gt sqrt((4)/(3))` |
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| 207. |
The ratio of molecular mass of two radioactive substances is `3//2` and the ratio of their decay cosntatnt is `4//3`. Then. The ratio of their initial activity per mole will beA. 2B. `(8)/(9)`C. `(4)/(3)`D. `(9)/(8)` |
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Answer» Correct Answer - C |
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| 208. |
An element `A` decays into element `C` by a two-step process : `A rarr B + ._2 He^4` `B rarr C + 2 e overline` Then.A. A and C are isotopesB. A and C ae isobarsC. B and C are isotopesD. A and B are isobars |
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Answer» Correct Answer - A |
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| 209. |
Assertion: During de-excitation from n=6 to n=3 , total six emission lines may be obtained. Reason: From `n=n to n =1 total (n(n-1)/(2)` emission lines are obtained.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason or true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A::B Let us take n=3 as N=1. Then, n=6 means N=4. So, total number of emission lines between N =1 and N =4 are `(N(N-1)/(2)) = (4xx3)/(2) =6.` |
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| 210. |
Assertion: Photoelectric effect proves the particle nature of light. Reason: Photoemission starts as soon as light is incident on the metal surface, provided frequency of incident light is greater than or equal to the threshold frequency.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason or true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A |
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| 211. |
If light of wavelength of maximum intensity emitted from surface at temperature `T_(1)` Is used to cause photoelectric emission from a metallic surface, the maximum kinetic energy of the emitted electron is 6 ev, which is 3 time the work function of the metallic surface. If light of wavelength of maximum intensity emitted from a surface at temperature `T_(2) (T_(2)=2T_(1))` is used, the maximum kinetic energy of the photo electrons emitted isA. 10eVB. 8eVC. 14eVD. 12eV |
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Answer» Correct Answer - C |
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| 212. |
A graph regarding photoelectric effect is shown between the maximum kinetic energy of electrons and the frequency of the incident light. On the basis of data as shown in the graph, calculate (a) threshold frequency , (b) work- function, (c ) planck constant |
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Answer» Correct Answer - A::B::C::D (a) `f_0 = f_A = 10xx10^(14)Hz = 10^(15) Hz` (b) `W=|K_(max)|_c = 4 eV` (c ) `W = hf_0 rArr h = (W)/(f_0).` |
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| 213. |
Using the known values for haydrogen atom, calculate. (a) radius of thirdj orbit for `Li^(+2)`. (b) speed of electron in fourth orbit for `He^+`. (c ) angular momentum of electron in 3rd orbit of `He^+.` |
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Answer» Correct Answer - A::B::C (a) `Z = 3 for Li^+2` . Further we know that `r_n = (n^2)/(Z) a_0` . Substituting, `n = 3, Z = 3` and `a_0 =0.529Å` . We have r_3 for `Li^(+2) = ((3)^2/(3)) (0.529)Å`. (b) `Z = 2 for He^+`. Also we know that . `v_n = (Z)/(n) v_1` .Substituting,` n = 4, Z=2 and V_1 = 2.19xx10^6 m//s` . We get, `v_4 for `He^+ = ((2)/(4)) ((2.19xx10^6))m//s` . `= 1.095xx10^6m//s`. (c ) `L_n = n((h)/(2pi))` . For `n = 3, L_3 = 3((h)/(2pi))` |
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| 214. |
A hydrogen atom and a `Li^(2+)` ion are both in the second excited state. If `l_H` and `l_(Li)` are their respective electronic angular momenta, and `E_H and E_(Li)` their respective energies, then (a) `l_H gt l_(Li) and |E_H| gt |E_(Li)|` (b) `l_H = l_(Li) and |E_H| lt |E_(Li)|` (C ) `l_H = l_(Li) and |E_H| gt |E_(Li)|` (d) `l_H lt l_(Li) and |E_H| lt|E_(Li)|` |
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Answer» Correct Answer - B in second excited state n=3 So, `l_H = l_(Li) = 3((h)/(2pi))` while `E prop Z^2 and Z_H = 1, Z_(Li) = 3` So, `|E_(Li)| = 9| E_H|` or `|E_H|lt|E_(Li)|.` |
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| 215. |
As an electron makes a transition from an excited state to the ground state of a hydrogen like atom/ion (a) kinetic energy, potential energy and total energy decrease (b) kinetic energy decreases, potential energy increases but total energy remains same (c ) kinetic energy and total energy decrease but potential energy increases (d) its kinetic energy increases but potential energy and total energy decrease |
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Answer» Correct Answer - D The expressions of kinetic energy, potential energy and total energy are `K_n = (me^4)/(8epsilon_0^2n^2h^2) rArr K_n prop (1)/(n^2)` `U_n = (-me^4)/(4epsilon_0^2n^2h^2) rArr `U_n prop - (1)/(n^2)` and `E_n = (-me^4)/(8epsilon_0^2n^2h^2) rArr E_n prop -(1)/(n^2)` in the transition from some excited state to ground state, the value of n decreases, therefore kinetic energy increases, but potential and total energy decrease. |
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| 216. |
Find variation of angular speed and time period of single. electron of hydrogen like atoms with n and Z. |
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Answer» Correct Answer - A::B::C Angular speed, `omega = (v)/(r )` . Now, `v prop (z)/(n)` and `r prop (n^2)/(Z)` . `:. Omega prop ((Z//n)/(n^2//Z))` or `omega prop (Z^2)/(n^3)` . Time period, `T=(2pi)/(omega)` or T prop(1)/(omega)` :. T prop (n^3)/(Z^2)`. |
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| 217. |
A metallic surface is illuminated alternatively with light of wavelenghts `3000 Å` and `6000 Å`. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1 . Calculate the work function of the metal and the maximum speed of the photoelectrons in two cases. |
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Answer» Correct Answer - A::C `E_1 = (12375)/(3000) = 4.125 eV` `E_2 = (12375)/(6000) =2.0625 eV` Maximum speed ratio is 3:1. Therefore, maximum kinetic ratio is 9:1. Now, `9K_(max) = 4.125 - W` `K_(max) = 2.0625 - W` Solving these two equations, we get `W ~~ 1.81 eV. and K_(max) = 0.26 eV` Putting `K_(max) = (1)/(2)mv_(max)^2` we can find `v_(max)`. Here, m is the mass of electron. |
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| 218. |
find kinetic energy, electrostatic potential energy and total. energy of single electron in 2nd excited state of `Li^(+2)` atom. |
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Answer» Correct Answer - A::B::C `:. E_I^H =0 13.6eV` Further `E prop (Z)^2)/(n^2), For `Li^2,`, Z =3` and for 2 end excited stat4e n =3 `E = -13.6 ((33)/(3))^2 =-13.6V` `K = |/E| = 13.6eV` U= 2E =-27.2eV` |
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| 219. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. (a) 6hB. (b) 12hC. (c) 24hD. (d) 128 h |
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Answer» Correct Answer - B From `R=R_0(1/2)^n` we have, `1=64(1/2)^n` or `n=6=number of half-lives` `:.` `t=nxxt_(t_1//2)=6xx2=12h` |
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| 220. |
In a `p-n` junction–A. new holes and conduction electrons are produced continuously throughout the materialB. new holes and conduction electrons are produced continuously throughout the material except in the depletion regionC. holes and conduction electrons recombine continuously throughout the materialD. holes and conduction electrons recombine continuously throughout the material except in the depletion region |
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Answer» Correct Answer - A |
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| 221. |
In an n- p-n transistor cirrect , the collectorcurrect is `10m A if 90 %` of the electrons reach the collector.A. the emitter currect will be `9mA`B. the base currect will be `1 mA`C. the emitter currect will be `11 mA`D. the base currect will be `- 1 mA` |
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Answer» Correct Answer - B::C `1_(e) = 10mA` `90%` of eklectrons emitted produce a cellector currect of `10mA` . The base currect `I_(b) = 10% of aU_(c) = (10)/(100) xx 10 = 1mA` Now I_(b) =I_(c) = 1 + 10 = 11mA` |
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| 222. |
Selected the correct statement from the followingA. A diode can be used as a neclifierB. A triode connot be used as a recifierC. the currect is a diode is always propotianal to the applilled vollageD. The linner porton odf the `1 - V` characteristic of a triode is used for amplification without distortion |
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Answer» Correct Answer - A A diode can be used as a rectifier. |
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| 223. |
…… bissing of `p-n` junction . Affers high resistance to currect flow across the junction . The biasing is obtained by connecting the `p - side` to the ……. Terminal of the battery |
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Answer» Correct Answer - A Reverse , negative terminal |
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| 224. |
for a give a plate voltage , the plate currect in a triode cvalve is maximum when the potential ofA. the grid is positive and plate is negetiveB. the grid is zero and plate is positiveC. the grid is negative and plate is positiveD. the grid is positive and plate is positive |
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Answer» Correct Answer - D in the currect option . The electrons by emitted are collected to the maximum by the in this case. |
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| 225. |
If the ratio of the concentration of electron to that of holes in a semiconductor is `(7)/(5)` and the ratio of currect is `(7)/(4)` then what is the ratio of their drift velocities ?A. `(5)/(8)`B. `(4)/(5)`C. `(5)/(4)`D. `(4)/(7)` |
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Answer» Correct Answer - C `I_(a))/(I_(b)) = (n_(e) edv_(e))/(n_(b) edv_(b)) rArr (7)/(4) = (7)/(5) xx (nu_(e))/(nu_(b)) rArr (nu_(e))/(nu_(b)) = (5)/(4)` |
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| 226. |
The circuit has two opposotively connected ideal diodes in parallel what is the currect flowing in the circuit ? A. `1.71 A `B. `2.00 A`C. `2.31 A`D. `1.33A` |
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Answer» Correct Answer - B `D_(2) ` is forward biased whereas `D_(1)` is reversed blased so effective of the circuit `R = 4 + 2 6 Omega` ` :. I = (12)/(6) = 2A ` |
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| 227. |
When a monochromatic point source of light is at a distance of `0.2` m from a photoelectron cell the cut off voltage and the saturation currect are respectively `0.6V` and `18.0 mA` if the same is plased `0.6 m` away from the photoelectric cell , thenA. the stopping potential will be `0.2Volt`B. the stopping potential will be `0.6 Volt`C. the stopping potential will be `6.0 Volt`D. the stopping potential will be `2.0 Volt` |
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Answer» Correct Answer - B::D Since the stopping potential depands on the frequency and not on the intensity and the source is same , the stopping potential remain unaffected . The sqatutation currect depends on the intensity of incident light on the cuthode of the photocell which in turn depends on the distance of the source from the cathode the intensity (i() of lioght is inversely propotional to the square of the distance between the light source and photcell `1 prop (1)/(r^(2))` and saturation currect `prop 1` `rArr Saturation Currect prop (1)/(r^(2))` `rArr ((Sataration Currect)_(fanal)/((Sataration Currect) _(mital) = (r_(initial)^(2))/(r_(final)^(2))` `rArr (Saturation Currect)_(final) = (0.2 xx 0.2)/(0.6 xx 0.6) xx 18 = 2mA` |
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| 228. |
A working transitior with its three legs marked `P, Q and R` is tested using a multimeter No conduction is found between `P, Q `by connecting the common (negative) terminal of the multimeter to `R` and the ofther (positive) terminal to or `Q` some resistance is seen on the multimeter . Which of the following is true for the transistor ?A. it is an npn transistor with `R` as baseB. it is a pop transistor with `R` as collectorC. it is a pop transistor with `R` as emitterD. it is an npn transistor with `R` as collector |
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Answer» Correct Answer - A It is a n- p- n transition with R as base |
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| 229. |
The diffusion current in a p-n junction isA. p-side to n-sideB. n-side to p-sideC. p-side to n-side if the junction is forward biased and in the opposite direction if it is reverse biasedD. n-side to p-side if the junction is forward biased and in the opposite direction if it is reverse based |
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Answer» Correct Answer - A |
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| 230. |
…… bissing of `p-n` junction . Affers high resistance to currect flow across the junction . The biasing is obtained by connecting the `p - side` to the ……. Terminal of the battery |
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Answer» Correct Answer - A Reverse , negative terminal |
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| 231. |
The electron condactivety of a samiconductor increases `2480 nm` is incident on it . The hand gap in (eV) for the semicondactor isA. `25 eV`B. `1.1 eV`C. `0.7 eV`D. `0.5 eV` |
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Answer» Correct Answer - D Band gap = energy of photon of wavelength `2480 nm` so, `delta E = (hc)/(lambda) = ((6.63 xx 10^(-34) xx 3 xx 10^(8))/(2480 xx 10^(-9))) = (1)/(1.6 xx 10^(-19)) eV` `= 0.5 eV` |
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| 232. |
The mean free path of a conduction electron in a metal is `5 xx 10^(-8) m`. The electric field, required to be applied across the conductor so as to impart `1eV` energy to the conduction electron , will beA. `1 xx 10^(-7) V//m`B. `2 xx 10^(7) V//m`C. `3 xx 10^(7) V//m`D. `4 xx 10^(7) V//m` |
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Answer» Correct Answer - A |
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| 233. |
Which of the following statements is correct ?A. when forward bias is applied on a `p-n` junction then current does not flow in the circuitB. rectification of alternating current can not be achieved by `p-n` junctionC. when reverse bias is applied on a `p-n` junction then it acts as a conductorD. some potential gap developed across the `p-n` junction when it is formed |
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Answer» Correct Answer - A |
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| 234. |
The decay constant of a radiaoacation sample is `lambda` . The half life and mean life of the sample are respectively given byA. `1// lambda and (in 2 )// lambda`B. ` (in 2 )// lambda`and 1// lambda `C. ` lambda (in 2 ) and 1// lambda `D. ` lambda //(in 2 ) and 1// lambda ` |
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Answer» Correct Answer - B `T_(1//2) = (ln 2 )/(lambda)` and Mean life `tau = (1)/(lambda) ` |
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| 235. |
The distance between consecutive maxima and minima is given by-A. `lambda//2`B. `2 lambda`C. `lamda`D. `lambda//4` |
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Answer» Correct Answer - D |
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| 236. |
Statement I : Penetration power of hard X-ray is more than that of soft X-ray. Statement II : Hard X-ray is used for engineering purpose while soft X-ray is used for medical purpose.A. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true. |
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Answer» Correct Answer - B |
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| 237. |
When a hydrogen atom emits a photon in going from n=5 to n=1, its recoil speed is almostA. 4 m/sB. 800 m/sC. 3mm/sD. `01. mm//s` |
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Answer» Correct Answer - A |
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| 238. |
STATEMENT - 1 If the accelerating potential in an X - rays tube is increased, the wavelength of the characteristic X- rays do not change .STATEMENT -2 When an electron beam strikes the target in an X- rays tube, part of the kinectic energy is converted into X - rays energy .A. 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 a NOT a correct explanation for Statement - 1C. Statement - 1 is true , Statement -2 is falseD. Statement - 1 is false , Statement -2 is true |
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Answer» Correct Answer - B (b) Statement 1 : The wavelength of characteristic X- rays depends on the of atome of which the targetmate meterial is mode , it does not depend on the accelerating potential therefore statement 1 is true Statement 2 : when an electric beam strikes the target in an X- rays tube , part of the kinetic energy is converted into X- ray energy .This statement is true But statement 2 does not explain statement 1 |
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| 239. |
In X-ray tube , when the accelerating voltage `V` is halved, the difference between the wavelength of `K_(alpha)` line and minimum wavelength of continuous X-ray spectrumA. remains constantB. becomes more than two timesC. becomes halfD. becomes less than two times |
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Answer» Correct Answer - D |
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| 240. |
`A overset(lambda)rarr B overset(2 lambda)rarr C` `T=0 ,N_(0) , 0 `, ` T N_(1) N_(2) N_(3)` The ratio of `N_(1)" to "N_(2) ` is maximum I s |
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Answer» Correct Answer - 2 |
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| 241. |
The ratio between acceleration of the electron in singley ionized helium atom and duby ionized lithium aomt (both in ground state) is (x/27). Find value of n. |
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Answer» Correct Answer - 8 |
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| 242. |
A deuteron and an `alpha` - partical have same kinetic energy. Find the ratio of their de-Broglie wavelengths. |
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Answer» Correct Answer - B `lambda = (h)/(sqrt(2Km) rArr lambda prop (1)/(sqrtm)` (lambda_b)/(lambda_alpha) = sqrt((m_alpha)/(m_d)) = sqrt((4)/(2)) = sqrt2.` |
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| 243. |
Find the ionisation energy of a doubly lonized lithium atom. |
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Answer» Correct Answer - A::B::D `Z = 3 for doubly ionized atom E prop Z^2` lonization energy of hydrogen atoms is 13.6 eV. `:.` lonisation energy of this atom `(3)^2(13.6) =122.4 eV |
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| 244. |
Find the de-Broglie wavelengths of (a) a 46 g golf ball with a velocity of 30m/s (b) an electron with a velocity of `10^7 m//s.` |
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Answer» Correct Answer - A::B::C::D `lambda = (h)/(mv)` (a) `lambda = (6.63xx10^(-34)/((46xx10^(-3)(30))` `= 4.8xx10^(-34)m` (b) `lambda = (6.63xx10^(-34))/(9.31xx10^(-31)xx10^7)` `=7.12xx10^(11) m.` |
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| 245. |
The electric field associated with a light wave is given by `E= E_0 sin [(1.57x 10^7 m^(-1)(x-ct)].` Find the stopping potential when this light is used in an experiment on photoelectric affect with a metal having work - function 1.9 eV. |
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Answer» Correct Answer - A::B ` f = (omega)/(2pi) = (1.57xx10^7)_c)/(2pi)` `E = (hf)/(1.6xx10^(-19) eV` `=((6.63xx10^(-34)(1.57xx10^7)(3xx10^8)/(1.6xx10^(-19)xx2xxpi)` =3.1 eV `K_(max) = E -W = 1.2 eV.` |
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| 246. |
if we assume only gravitational attraction between proton and electron in hydrogen atom and the Bohr quantizaton rule to be followed, then the expression for the ground state energy of the atom will be (the mass of proton is M and that of electron is m.)A. increases 4 timesB. decreases 4 timesC. increases 8 timesD. decreases 8 times |
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Answer» Correct Answer - B `(mv^2)/(r ) = (GMm)/(r^2)…..(i)` `mvr =(h)/(2pi) (for n =1) ….(ii)` Solving these two equations, we can find v and r. Then, `E = (1)/(2) mv^2 - (GMM)/(r )` |
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| 247. |
The half - life period of a radioactive element X is same as the mean - life time of the another radicoactive electront Y initial both of then the same number of atom . ThenA. X and Y have the same decay rate initial lyB. X and Y dacay at the same decay rate alwaysC. Y will dacay at a faster rate then XD. X will dacay at a faster rate then Y |
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Answer» Correct Answer - C `(1_(1//2)) s = (1_(mean))Y rArr (0.693) (lambda_(z)) = (1)/(lambda_(Y))` `:. Lambda_(x) = 0.693 lambda_(Y)` `lambda_(x) lt lambda_(Y) `Now ratye of dacay `= lambdaN` initialy , number of atoms (N) of both are equal but since l`lambda_(Y) lt lambda_(x) `therefore Y will decay at a faster rate from x |
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| 248. |
The work function of a substance is `4.0 eV ` The longest wavwlength of light that can cause photoelectron emission from this substance is approximatelyA. `540 nm`B. `400 nm`C. `310 nm`D. `220 nm` |
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Answer» Correct Answer - C `lambda_( min) = (hc)/(W) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(4(1.6 xx 10^(-19))) = 310 xx 10^(-9)m ` ` = 310 nm` |
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| 249. |
Consider the following nuclear reaction, `X^200rarrA^110+B^90+En ergy` If the binding energy per nucleon for X, A and B are `7.4 MeV`, `8.2 MeV` and `8.2 MeV` respectively, the energy released will beA. 200 MeVB. 160 MeVC. 110 MeVD. 90 MeV |
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Answer» Correct Answer - B |
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| 250. |
Suppose an electron is attracted toward the origin by a force`(k)/(r )` where `k` is a constant and `r` is the distance of the electron from the origein .By appling Bohr model to this system the radius of the `n^(th)` orbital of the electron is found to be `r_(n)` and the kinetic energy of the electron to be `T_(n)` , Then which of the following is true ?A. `T_(n) prop (1)/(n^(2)) , r_(n) prop n^(2)`B. `T_(n)` independent of `n , r_(n) prop n `C. `T_(n) prop (1)/(n) , r_(n) prop n`D. `T_(n) prop (1)/(n) , r_(n) prop n^(2)` |
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Answer» Correct Answer - B When `F= (k)/(r ) = contripetal force , then ( (k)/(r ) = (m v^(2))/(r )` `rArr m v^(2) rArr kinetic energy is constant rArr T` is independent of `n` |
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