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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.
| 1151. |
The work function `(phi)` of some metals is listed below . The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is : |
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Answer» Correct Answer - 4 |
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| 1152. |
Visible light consists of rays with wavelengths in the approximate range of:A. 4000 Ã… to 7500 Ã…B. `4xx10^(-3)cm` to `7.5xx10^(-4)cm`C. 4000 nm to 7500 nmD. `4xx10^(-5)m` to `7.5xx10^(-6)m` |
| Answer» Correct Answer - A | |
| 1153. |
Which among the following pairs are having different number of toatl electrons ?A. `Na^(+)and Al^(+3)`B. `P^(-3)and Ar`C. `Mg^(+2) and Ar`D. `O^(-2) and F^(-)` |
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Answer» The electronic configuration of `Mg^(+2) is 2,8` `therefore` Total no. of electrons `=10` The electronic configuration of Ar is 2,8,8 `therefore` Total no. of electrons `=18` Hence, `Mg^(+2)` and Ar are having different number of total electrons. |
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| 1154. |
The frequency of radiations emitted when electron falls from `n = 4` to `n = 1` in `H-"atom"` would be (Given `E_1` for `H = 2.18 xx 10^-18 J "atom"^-1` and `h = 6.625 xx 10^-34 Js`.)A. `1.54 xx 1015 s^-1`B. `1.03 xx 1015 s^-1`C. `3.08 xx 1015 s^-1`D. `2.0 xx 1015 s^-1` |
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Answer» Correct Answer - C ( c) Ionisation energy of `H = 2.18 xx 10^-18 J "atom"^-1` :. `E_1` (energy of `Ist` orbit of `H-"atom"`) =` -2.18 xx 10^-18 J "atom"^-1` :. `E_n = (-2.18 xx 10^-18)/(n^2) J atm^-1` `Z = 1` for `H - "atom"` `Delta E = E_n - E_1 = (-2.18 xx 10^-18)/(4^2) - (-2.18 xx 10^-18)/(1^2)` `Delta E = -2.18 xx 10^-18 xx (-15)/(16)` =`+ 2.0437 xx 10^-18 J at om^-1` `v = (Delta E)/(h) = (2.04 xx 10^-18)/(6.626 xx 10^-34)` =`3.084 xx 10^15 s^-1 at om^-1`. |
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| 1155. |
Ground state energy of H-atom is `(-E_(1))`,t he velocity of photoelectrons emitted when photon of energy `E_(2)` strikes stationary `Li^(2+)` ion in ground state will be:A. `v = sqrt((2(E_(p)-E))/(m))`B. `v =sqrt((2(E_(p)+9E))/(m))`C. `v = sqrt((2(E_(p)-9E))/(m))`D. `v = sqrt((2(E_(p)-3E))/(m))` |
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Answer» Correct Answer - C `E_(0) =E_(0) +(1)/(2)mv^(2) rArr v = sqrt((2(E_(p)-E_(0)))/(m))` But `E_(0) = 9E` |
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| 1156. |
Which of the following conditions is incorrect for a well behaved wave function `(Phi)`?A. `psi` must be finiteB. `psi` must be single valuedC. `psi` must be infiniteD. `psi` must be continuous |
| Answer» Correct Answer - C | |
| 1157. |
The correct set of four quantum number for the valence (outermost) electron of radiation `(Z = 37)` isA. `5,0,0, +(1)/(2)`B. `5,1,0,+(1)/(2)`C. `5,1,1,+(1)/(2)`D. `5,0,1,+(1)/(2)` |
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Answer» Correct Answer - A |
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| 1158. |
The four quantum number for the valence shell electron or last electron of sodium (Z=11) isA. n=2, l=1, m=-1, `s=-1/2`B. n=3, l=0, m=0, `s=-+ 1/2`C. n=3, l=2, m=-2, `s=-1/2`D. n=3, l=2, m=2, `s=+ 1/2` |
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Answer» Correct Answer - B `Na_11=2,8,1=1s^2, 2s^2 2p^6, 3s^1` n=3, l=0, m=0, s=+1/2 |
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| 1159. |
The values of four quantum number of valence electron of an element are `n = 4, l = 0, m = 0 and s=+(1)/(2)` . The element is:A. KB. TiC. NaD. Sc |
| Answer» Correct Answer - A | |
| 1160. |
The energies of orbitals in hydrogen -like spectries depend on the quantum number (s) ……. |
| Answer» Correct Answer - n | |
| 1161. |
The energies of orbitals in a multi -electron atom depend on the quantum number (s) ……… |
| Answer» Correct Answer - n and l | |
| 1162. |
Which of the following statements is not correct ?A. The shape of an atomic orbital depends on the azimuthal quantum numberB. The orientation of an atomic is given by magnetic quantum numberC. The energy of an electron in an atomic orbital of multi electron atom depends on the principal quantum number onlyD. The number of degenerate atomic orbitals of one type depends on the values of azimuthal and magnetic quantum numbers. |
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Answer» Correct Answer - C ( c) For multi electron atom energy depends on `n` and `l (n + l)` rule. |
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| 1163. |
If no. of protons in `X^(-2)` is `16`. Then no. of `e^(-)` in `X^(+2)` will be-A. `14`B. `16`C. `18`D. None |
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Answer» `because` No. of proton in `X^(-2)` is `=16` `therefore` No. of electron in `X^(+2)` is `=14` |
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| 1164. |
Which of the following statements is not correct ?A. Special stability of half-filled and fully-filled atomic configurations amongst `s-` and p-block elements is reflected in ionization potential tends along a period.B. Special stability of half-filled and fully-filled atomic configurations amongest `s-` and p-block elements is reflected in electron affinity trends along a period.C. Aufbau order is not obeyed in cases where energy difference between `ns` and `(n - 1) d` subshell is large.D. Specical stability of half-filled subshell is attributed to higher exchange energy of stabilization. |
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Answer» Correct Answer - C ( c) Arfbau order is obeyed in cases where energy difference between `ns` and `(n - 1) d` subshell is large. |
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| 1165. |
In `C^(12)` atom if mass of `e^(-)` is doubled and mass of proton is halved, then calculate the percentage change in mass no. of `C^(12)` |
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Answer» `{:(,,._(6)C^(12),,),(,e^(-),P^(+),N^(o),),(,6,6,6,Ararr12),(,12,3,6,Ararr9):}` `%` change `=(3)/(12)xx100=25%` |
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| 1166. |
Which of the following is true ?A. The outer electronic configuration of the ground state chromium atom is `3d^4 4 s^2`.B. Gamma rays are electromagnetic radiations of wavelength of `10^-6 cm` to `10^-5 cm`.C. The energy of the electron in the `3d` orbital is less than that in the `4s` orbital of a hydrogen atom.D. The electron density in the `xy-"plane"` in `3 d_(x^2 - y^2)` orbital is zero. |
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Answer» Correct Answer - C ( c) (a) False. The configuration is `3 d^5 4 s^1`. (b) False. The wavelength of gamma rays is of the order of `10-11 m`. ( c) True. In hydrogen atom, the energy of an electron depends only on the principal quantum number of the orbital which it occupies. (d) False. In xz-plane, there is no electron density if an electron occupies `3 d_(x^2 - y^2)` orbital. |
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| 1167. |
An `alpha`-particle having `K.E. =7.7MeV` is scattered by gold `(z=79)` nucleus through `180^(@)` Find distance of closet approach. |
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Answer» `K.E.=7.7MeV=7.7xx10^(6)xx1.6xx10^(-19)J=1.23xx10^(-12)J` `(1)/(4piepsilon_(0))=9xx10^(9)Nm^(2)C^(-2)` Using we get `: (9xx10^(9)xx2xx79xx(1.6xx10^(-19))^(2))/(1.23xx10^(-12))` `r_(0)=3xx10^(-14)m` From the above example it is clear that nuclear diamension cannot be greater than `3xx10^(-14)m` |
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| 1168. |
An electron has a speed of `40m//s`, accurate up `99.99%`.What is the uncertainty in locating position ? |
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Answer» Given, `Delta v = (0.01)/(100) xx 40 = 0.004 ms^(-1)` we know, `Deltax.Delta v ge (h)/(4pim)` `Delta x ge (h)/(4pi mDelta v)` `Delta x = (6.625 xx 10^(-34))/(4 xx 3.14 xx 9.1 xx 10^(-31) xx 0.004)` `Delta x = 1.45 xx 10^(-2)m` |
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| 1169. |
A quantum of light having energy E has wavelength equal to `7200 A^(@)`. The frequency of light which corresponds to energy equal to 3E, isA. `1.25 xx 10^(14)s^(-1)`B. `1.25 xx 10^(15)s^(-1)`C. `1.25 xx 10^(13)s^(-1)`D. `1.25 xx 10^(14)s^(-1)` |
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Answer» Correct Answer - B `v_(1) = (c)/(lambda_(1)), v_(2) = (c)/(lambda_(2)),(E_(1))/(E_(2)) = (v_(1))/(v_(2))` |
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| 1170. |
Electronic transition in `He^(+)` ion takes from `n_(2) " to " n_(1)` shell such that : `2n_(2)+3n_(1)=18` `2n_(2)+3n_(1)=6` What will be the total number of photons emitted when electrons transit to `n_(1)` shell?A. 21B. 15C. 20D. 10 |
| Answer» Correct Answer - D | |
| 1171. |
The minimum kinetic energy of a ground state hydrogen atom required to have head-on collision with another ground state hydrogen atom but at rest to produce a photon is given by |
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Answer» Correct Answer - 20.4 eV |
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| 1172. |
First line of Lyman series of hydrogen atom occurs at `lamda=x`Ã…. The corresponding line of `He^(+)` will occur at:A. 4xB. 3xC. x/3D. x/4 |
| Answer» Correct Answer - D | |
| 1173. |
Show that de Broglie wavelength of electrons accelerated V volt is very nearly given by: `lamda("in Ã…")=((150)/(V))^(1//2)` |
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Answer» `lamda=(h)/(sqrt(2eVm))` `lamda=[(h^(2))/(2eVm)xx10^(20)]^(1//2)`Ã… `=[((6.626xx10^(-34))^(2)xx10^(20))/(2xx1.6xx10^(-19)xxVxx9.1xx10^(-31))]^(1//2)=[(150)/(V)]^(1//2)` |
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| 1174. |
How many quantum number are needed in designate an orbital ? Name them |
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Answer» Correct Answer - C Three principal quantum number , azimuthal quantum number, and magnetic quantum number |
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| 1175. |
The critical wavelength for producing photoelectric effect in a metal is `2500 Å` What wavelength would be nuccesary be produce photoelectric effect from this metal , having twice the KE of these produced at `2000 Å` |
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Answer» `KE_(2000) = hv_(2000) - hv_(0)= hc ((1)/(lambda_(2000)) - (1)/(lambda_(0)))` `KE_("new") - hv_("new") = hv_(0)= hc ((1)/(lambda_("new")) - (1)/(lambda_(0)))` KE_("new") = 2 xx KE_(2000)` `2= (hc((1)/(lambda_("new")) - (1)/(2500)))/(hc((1)/(2000) - (1)/(2500))) , lambda = 1666.66 Å` |
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| 1176. |
Which of the following transition will have a wavelength different than that observed in rest of the transitions ?A. H-atoms , transition from 3rd level to 1st level .B. `He^(1+)` ion , transition from 5th excited state to 1st excited state .C. `Li^(2+)` ion , transition from 9th level to 3rd level.D. `Be^(+3)` ion, transition from 11th excited state to 3rd level . |
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Answer» Correct Answer - D |
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| 1177. |
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n=4` to `n=2` of `He^(+)` spectrum ?A. n=4 to n=3B. n=3 to n=2C. n=4 to n=2D. n=2 to n=1 |
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Answer» Correct Answer - D For `He^+` ion, `1/lambda=Z^2R[1/n_1^2-1/n_2^2]` `(2)^2 R[1/2^2 -1/4^2] ="3R"/4` For hydrogen atom , `1/lambda=R[1/n_1^2-1/n_2^2]` `"3R"/4=R[1/n_1^2-1/n_2^2]` or `1/n_1^2-1/n_2^2=3/4` `n_1=1` and `n_2=2` |
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| 1178. |
Which of the following transitions have minimum wavelengthA. `n_4 to n_1`B. `n_2 to n_1`C. `n_4 to n_2`D. `n_3 to n_1` |
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Answer» Correct Answer - A For `n_4 to n_1` , greater transition, greater the energy difference, lesser will be the wavelength . |
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| 1179. |
1 mole of photon, each of frequency `2500 s^(-1)` , would have approximately a total energy ofA. 10 ergB. 1 JouleC. 1 eVD. 1 MeV |
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Answer» Correct Answer - A `upsilon=2500 s^(-1)` `therefore` Energy of a photon , E=hv `=(6.626xx10^(-34)xx2500)`J `because` Energy of one mole of photon `=(6.626xx10^(-34)xx2500xx6.023xx10^23)J` `=9.977xx10^(-7) J ~~ 10` erg |
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| 1180. |
Based on equation `E = -2.178 xx 10^-18 J((Z^2)/(n^2))`, certain conclusions are written. Which of them is not correct ?A. For n=1, the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbitB. The negative sign in equation simply means that the energy of electron bound to nucleus is lower than it would be if the electrons were at the infinite distance from the nucleusC. larger the value of n, the larger is the orbit radiusD. Equation can be used to calculate the change in energy when the electron change orbit |
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Answer» Correct Answer - A Correct answer is - for n=1 the electron has more negative energy than it does for n=6 which means that the electron is less loosely bound in the smallest allowed orbit |
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| 1181. |
Total no of lines in Lyman series of H spectrum will be- (where n=no. of orbits)A. `n`B. `n-1`C. `n-2`D. `n(n+1)` |
| Answer» Correct Answer - B | |
| 1182. |
Difference in wavelength of two extreme lines of Lyman series in emission spectrum of `He^(+)` would be :A. `(1)/(12R_(H))`B. `(12)/(R_(H))`C. `(1)/(4R_(H))`D. `(1)/(3R_(H))` |
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Answer» Correct Answer - A |
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| 1183. |
Two hydrogen atom collide Collide head on and end up with zero kinetic energy .Each atom then emit a photon of wavelength `121.6 nm` .Which transition leads to the wavelength ? How fast werte the hydrogen atom transition before collision ? |
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Answer» Wavelength is emitted in UV regain and thus `n_(1) = 1`1 For H atom `(1)/(lambda) = R_(H) [(1)/(1^(2)) - (1)/(n^(2))]` `:. -(1)/(121.6 xx 10^(-6)) = 1.097 xx 10^(7)[(1)/(1^(2))- (1)/(n^(2))]` `:.n = 2` Also the energy released is due to collision and all the kinetic energy is released in the form of photon `:.(1)/(2) mv^(2) = (hv)/(lambda)` `(1)/(2) xx 1.67 xx 10^(-27) xx v^(2) = (6.625 xx 10^(-34) xx 3 xx 10^(8))/(121.6 xx 10^(-6))` `: , v = 4.43 xx 10^(4) m s^(-1)` |
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| 1184. |
Calculate the momentum of radiation of wavelength `0.33 nm`A. `2 xx 10^(-24)`B. `2 xx 10^(-12)`C. `2 xx 10^(-6)`D. `2 xx 10^(-48)` |
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Answer» Correct Answer - A `p = (h)/(lambda)` |
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| 1185. |
An ion has a charge of -1 . It has eighteen electrons and twenty neutrons. Its mass number isA. 17B. 37C. 18D. 38 |
| Answer» Correct Answer - B | |
| 1186. |
Bohr model cannot explain spectrum ofA. the hydrogen atom onlyB. all elementsC. any atomic or ionic species having one electron onlyD. the hydrogen molecule |
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Answer» Correct Answer - B (b) Bohr model cannot explain the spectrum of any species having multi electrons. It can explain only the spectrum of hydorgen and hydrogen like species having single electron only. |
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| 1187. |
Bohr model can be applied on :A. H atomB. `He^(+) "ion"`C. `H^(-) "ion"`D. `Li^(2+) "ion"` |
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Answer» Correct Answer - a b d |
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| 1188. |
Energy of H-atom in the ground state is `-13.6eV`. Hence energy in the second excited state isA. `-6.8 eV`B. `-3.4` eVC. `-1.51` eVD. `-4.53` eV |
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Answer» Correct Answer - C |
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| 1189. |
Select correct statement(s) about photoelectric effect.A. If green light ejects photoelectron from metal M then red light can never eject photoelectrons from metal M.B. If green light ejects photoelectron from metal M then red light may not eject photoelectrons even if it has high intensity for metal M.C. If green light ejects photoelectron from metal M with low photocurrent then more photo intense blue light must have high photocurrent for metal M.D. If green light ejects photoelectron then blue light must eject photoelectron with more energy for metal M. |
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Answer» Correct Answer - b c d |
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| 1190. |
The orbit having Bohr radius equal to 1st Bohr orbit of H-atom is :A. `n = 2 "of" He^(+)`B. n= 2 of `B^(+4)`C. n = 3 of `Li^(2+)`D. n = 2 of `Be^(+3)` |
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Answer» Correct Answer - D |
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| 1191. |
The orbital angular momentum of an electron is `sqrt3(h)/(pi).` Which of the following should not be the permissible value of orbit angular momentum of this electron revolving in Bohr orbit ?A. `h/(2pi)`B. `h/pi`C. `3.(h)/(2pi)`D. `2.(h)/(pi)` |
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Answer» Correct Answer - a b c |
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| 1192. |
In photo electric effect, if the energy required to over come the attractive forces on the electron (work function) of Li, Na and Rb are `2.41eV, 2.3eV` and `2.09eV` respectively, the work function of 'K' could approximately by in eVA. `2.52`B. `2.2`C. `2.35`D. `2.01` |
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Answer» Correct Answer - B As the size of atom increases, energy required to over come the attractive forces on the outer most electron decreases. |
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| 1193. |
The two electron have the following sets of quantum number X `3,2 -2,+ 1//2` `3,0,0, + 1//2` What is true of the followingA. X and Y have same energyB. X and Y have unequal energyC. X and Y have represent same electronsameD. None of the statement is correct |
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Answer» Correct Answer - B In `X, l = s ` In `Y,l = s` Energy of `d lt es` |
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| 1194. |
The number of electrons and neutrons of an element is 18 and 20 respectively . Its mass number isA. 17B. 37C. 2D. 38 |
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Answer» Correct Answer - D No. of proton and no. of electron =18 and No. of neutron =20 Mass number = P + N =18+20 =38 |
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| 1195. |
The angular momentum of an electron due to its spin is given as ……….. |
| Answer» `sqrt(s(s + 1)) ((h)/(2pi))` | |
| 1196. |
Number of unpaired electrons in inert gas isA. ZeroB. 8C. 4D. 18 |
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Answer» Correct Answer - A Number of unpaired electrons in inert gas is zero because they have full filled orbitals. |
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| 1197. |
Select the correct statements about the wave function `psi`.A. `psi` need not be realB. `psi` must be single valued, continuousC. `psi` has no physical significanceD. `psi^(2)` gives the probability density of finding the electrons |
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Answer» Correct Answer - A::B::C::D Properties of `Psi` |
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| 1198. |
The nucleus of the element having atomic number 25 and atomic weight 55 will containA. 25 protons and 30 neutronsB. 25 neutrons and 30 protonsC. 55 protonsD. 55 neutrons |
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Answer» Correct Answer - A No. of protons = Atomic no . =25 and no. of neutrons =55-25=30 |
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| 1199. |
In the Schrodingers wave equation `spi` repressentsA. OrbitB. Wave functionC. WaveD. Radial probability |
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Answer» Correct Answer - B `spi = `wave function |
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| 1200. |
Photo electric effect is not observed in case ofA. PotassiumB. RubidiumC. MagnesiumD. Cesium |
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Answer» Correct Answer - C Highly electropositive metals show photo electric effect. |
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