InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2201. |
The potential energy of the electron present in the ground state of ` Li^(2+)` ion is represented by :A. `+ (3e^(2))/(4 pi in_(0) r)`B. `-(3e)/(4pi in_(0)r)`C. `-(3e^(2))/(4pi in_(0) r^(2))`D. `-(3e^(2))/(4 pi in_(0)r)` |
|
Answer» Correct Answer - D |
|
| 2202. |
A photon of energy hv is absorbed by a free electron of a metal having work- function `phi lt hv`. Choose the correct option(s).A. Number of electrons coming out depends on magnitude of v .B. The electron is sure to come out with a kinetic energy `hv - phi`C. Either the electron does not come out or it comes out with a kinetic energy `hv - phi`D. It may come out with a kinetic energy less than and equal to `hv - phi` |
|
Answer» Correct Answer - D |
|
| 2203. |
What is the energy of photons with a wavelength of 434 nm?A. `2 . 76 xx 10^(5)` kJ//moleB. `2.76 xx 10^(2) "kJ"//"mole"`C. `2.76 xx 10^(-1) "kJ"//"mole"`D. `2.76 xx 10^(-4) "kJ"//"mole"` |
|
Answer» Correct Answer - B |
|
| 2204. |
At temperature T , the average kinetic energy of any particles is `(3)/(2)` kT. The de Broglie wavelength follows the order :A. Visible photon `gt` Thermal electron `gt` Thermal neutronB. Thermal proton `gt` Thermal electron `gt` Visible photonC. Visible photon `gt` Thermal neutron `gt` Thermal electronD. Thermal proton `gt` Visible photon `gt` Thermal electron |
|
Answer» Correct Answer - A |
|
| 2205. |
Energy of an electron is givem by `E = - 2.178 xx 10^-18 J ((Z^2)/(n^2))`. Wavelength of light required to excited an electron in an hydrogen atom from level `n = 1` to `n = 2` will be `(h = 6.62 xx 10^-34 Js` and `c = 3.0 xx 10^8 ms^-1`).A. `1.214 xx 10^(-7)m`B. `2.816 xx 10^(-7)m`C. `6.500 xx 10^(-7)m`D. `8.500 xx 10^(-7)m` |
|
Answer» Correct Answer - A |
|
| 2206. |
Which of the following is false?a.The energy of an electron in an orbital of a hydrogen-like species depends only on the principle quantum number n.b. The angular momentum of electron in an orbital of a multielectron atom depends on the quantum number l and mc. The experimental value of angular momentum of an orbital is given as `sqrt(l(l - 1)) ((h)/(2pi))`d. The z-component of angular momentum of an electron in an orbital is given as `m ((h)/(12pi))`A. The energy of an electron in an orbital of a hydrogen like species depends only on the principle quantum number `n`.B. The angular momentum of an electron in an orbital of a multielectron atom depends on the quantum numbers `l` and `m`.C. The expression of angular momentum of an electron in an orbital is given as `sqrt(l(l - 1)) ((h)/(2 pi))`.D. The z-component of angular momentum of an electron in an orbital is given as `m((h)/(2 pi))`. |
|
Answer» Correct Answer - B (b) (a) True. (b) False. The angular momentum depends only on the azimuthal quantum number. ( c) True. (d) True. |
|
| 2207. |
Assuming that atomic weight of `C^(12)` is `150` unit from atmic table, then according to this assumption, the weight of `O^(16)` will be:- |
| Answer» `{:(,therefore,12"amu"=150),(,therefore,1"amu"=(150)/(12)),(,therefore,16"amu"=(150)/(12)xx16=200"Unit"):}` | |
| 2208. |
The ratio of ionization energy of `H` and `Be^(+3)` is.A. `1 : 1`B. `1 : 3`C. `1 : 9`D. `1 : 16` |
|
Answer» Correct Answer - D (d) `IE = E_(prop) - E_n` `IE prop Z_2` So, `(IE_H)/(IE_(Be^(_3))) = (1)/(16)` `:. "ratio is" 1 : 16`. |
|
| 2209. |
The electron in ` Li^(2+)` ions are exctied form gound stte by absorbing ` 8. 4375Rh` energy /electron . How much emission lines are expected during de-excitation of electrons to ground state ? |
|
Answer» Correct Answer - 6 `:. Delta = 8. 4375Rh, E_(1Li^(2+)) =- Rh xx 9` and ` E_(4Li)^(2+) = ( -Rh xx 9)/(16) ` ` :. E_1 - E_4 = ( 9Rh)/1 - ( 9Rh)/(16) = ( 135Rh)/(16) = 8. 42375 Rh` ltbRgt Thus , de-excitaition will lead form ` 4^(th)` to 1st orbit . No. of spectral lines ltbRgt ` = sum Delta n =sum ( 4 -1) = sum 3 = 1+2+3 =6`. |
|
| 2210. |
Which of the following statementregarding cathode rays is not currect ?A. Cathode rays originate from the cathodeB. The charge and mass of the particle constituting cathode rays depends upon the the nature of the gasC. The charge and mass of the particles present does not depend upon the the metarial of the cathodeD. The charge and mass of the particle is much greater than that of anode rays |
|
Answer» Correct Answer - The charge and mass of the particle constituting cathode rays depends upon the the nature of the gas The charge and mass of the particle constituting cathode rays depends upon the the nature of the gas |
|
| 2211. |
The electronic configuration of gadolinium (Atomic number 64) isA. [Xe] `4f^(8), 5d^(9), 6s^(2)`B. [Xe] `4f^(7), 5d^(1), 6s^(2)`C. [Xe] `4f^(6), 5d^(2), 6s^(2)`D. [Xe] `4f^(3), 5d^(5), 6s^(2)` |
|
Answer» Correct Answer - B (b) `Gd_(64)=1s^(2),2s^(2)2p^(6),3s^(2)3p^(6)3d^(10), 4s^(2)4p^(6)4d^(10)4f^(7),5s^(2)5p^(6)5d^(1),6s^(2)` `=[Xe]4f^(7),5d^(1),6s^(2)` |
|
| 2212. |
Which of the following represents the electronic configuration of an element with atomic number 17A. `1s^2, 2s^2 2p^6 , 3s^1 3p^6`B. `1s^2 , 2s^2 2p^6 , 3s^2 3p^4 , 4s^1`C. `1s^2 , 2s^2 2p^6 , 3s^2 3p^5`D. `1s^2 , 2s^2 2p^6, 3s^1 3p^4 , 4s^2` |
|
Answer» Correct Answer - C Atomic no. 17 is of chlorine |
|
| 2213. |
An element has atomic number 37 the electronic configuration of the element isA. `(2 8)3s^(2)3p^(6) 3d^(10)4s^(2)5s^(6)4p^(5)`B. `(2 8)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)5s^(1)`C. `(2 8)3s^(2)3p^(6) 4s^(2)3d^(9)5s^(1)4p^(5)`D. None of the above |
|
Answer» Correct Answer - B `Rb=(Z=37),(2,8),3s^(2)3p^(6)3d^(6)3d^(10),4s^(2)4p^(6)5s^(1)` |
|
| 2214. |
The minimum angular momentum of an electron with the magnetic quantum numbers `-1,0,+1`A. `sqrt((3)/(2)(h)/(pi))`B. `(h)/(pi)`C. `(2h)/(pi)`D. `(3)/(2)(h)/(pi)` |
|
Answer» Correct Answer - B `l = 1` |
|
| 2215. |
The wave function, `psi_(n), l, m_(l)` is a mathematical function whose value depends upon spherical polar coordinates `(r,theta,phi)` of the electron and characterised by the quantum number `n,l` and `m_(l)`. Here `r` is distance from nucleus, `theta` is colatitude and `phi` is azimuth. In the mathematical functions given in the Table, `Z` is atomic number and `a_(0)` is Bohr radius. For the given orbital in Column I, the Only CORRECT combination for any hydrogen-like species isA. (II)(ii)(P)B. (I)(ii)(S)C. (IV)(iv)(R)D. (III)(iii)(P) |
|
Answer» Correct Answer - A |
|
| 2216. |
Statnment -I : Radial part of wave function for 2s & 2p is not same Statement -II : Radial part of wave function depends upon is zero at nucleus for all p & d orbitals. upon "l" and "m" only.A. If both Statement -I & Statement -II are True & the Statement -II is a correct explanation of the Statement I.B. If both Statement -I & Statement -II are True but the Statement -II is not a correct explanation of the Statement- I.C. If Statement -I is True but the Statement -II is False.D. If Statement -I is True but the Statement -II is True . |
|
Answer» Correct Answer - C Satement-1: Radial part of wave function is not same for two different orbitals. Statement -2: Radial part depend upon n and l bhot |
|
| 2217. |
The electrons, identified by quantum numbers n and l (i) `n=4l=1` (ii) `n=4l=0` (iii) `n=3l=2` (iv) `n=3l=1` can be placed in increasing order of energy from the lowest to highest asA. `(iv)lt(ii)lt(iii)lt(iv)`B. `(ii)lt(iv)lt(i)lt(iii)`C. `(i)lt(iii)lt(ii)lt(iv)`D. `(iii)lt(i)lt(iv)lt(ii)` |
|
Answer» Correct Answer - A (i) 4p (ii)4s (iii)3d (iv)3p |
|
| 2218. |
The wave function `Phi_(n,l,m_(1))` is a methematical function whose value depends upon shperical upon spherical polar coordinates `(r,theta,phi)` of the electron and characterized by the quantum numbers, n,l and `m_(1)`. Here r is distance from nucleus, `theta` is colatitude and `phi` is azimuth. in the mathematical functions given in the table, Z is atomic number, `a_(0)` is Bohr radius. . Q. For `He^(+)` ion, the only INCORRECT combination is:A. (II) (ii) (Q)B. (I) (i) (P)C. (II) (i) (Q)D. (I) (i) (S) |
| Answer» Correct Answer - D | |
| 2219. |
The angular wave function depends upon quantum numbers.A. n and lB. l and mC. l and sD. m and s |
|
Answer» Correct Answer - B only l and m |
|
| 2220. |
`e//m` ratio in case of anode ray experiment is different for different gases. The ion of gases formed after the ejection of electron are different of gas is different.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
|
Answer» Correct Answer - A (a) `e//m` ratio for particles in anode rays is different for different gases as different produce different positively charged particles upon ionisation. |
|
| 2221. |
What will be the uncertainly in velocity of a cricket ball of 100 g if the uncertainly in its position is `1.65 Å` ?A. `(10^(-23))/pi ms^(-1)`B. `(6.6)/pi 10^(-45)ms^(-1)`C. `4.65xx10^(-43)ms^(-1)`D. `(10^(-26))/pi ms^(-1)` |
|
Answer» Correct Answer - A According to Heisenberg uncertainty principle, `Deltax xxmDeltav=h/(4pi)` `1.65xx10^(-10)xx100/1000xxDeltav=(6.625xx10^(-34))/(4pi)` `Deltav=(6.625xx10^(-34+10))/(1.65xx4pi)xx10=(10^(-23))/pi m//s` |
|
| 2222. |
The modern atomic weight scale is based onA. `C^12`B. `O^16`C. `H^1`D. `C^13` |
| Answer» Correct Answer - A | |
| 2223. |
The charge/mass (e/m) ratio is maximum forA. protonB. neutronC. electronD. positron |
| Answer» Correct Answer - C | |
| 2224. |
In which one of the following , the number of protons is greater than neutrons but number of protons is less than the number of electronsA. `D_3O^+`B. `SO_2`C. `H_2O`D. `OH^(-)` |
| Answer» Correct Answer - D | |
| 2225. |
The number of neutrons in an atom of `""_(11)^(23)A` isA. 23B. 11C. 34D. 12 |
| Answer» Correct Answer - D | |
| 2226. |
Which of the followig sepcies has less number of protons than the number of neutrons?A. `._(6)^(12)C`B. `._(9)^(19)F`C. `._(11)^(23)Na`D. `._(12)^(24)Mg` |
| Answer» Correct Answer - B::C | |
| 2227. |
When a gas at reduced pressure is subjected to an electric discharge, the rays originating from the negative electrode areA. anode raysB. cathode raysC. X-raysD. radioactive rays |
| Answer» Correct Answer - B | |
| 2228. |
What maximum wavelength is required for a photon to dissociate : `X_(2)(g) rarr 2X(g)` if bond energy of `X-X` bond is `240kJ//mol` ? |
|
Answer» One photon breaks one bond only Energy of break `N_(A)` bonds `=240xx10^(3)J` Energy of break `1` bond `=(240xx10^(3))/(N_A)J` `implies(240xx10^(3))/(N_A)= (hc)/(lambda_(max))=(6.626xx10^(-34)xx3xx10^(8))/(lambda_(max))=lambda_(max)=4966 Å` |
|
| 2229. |
The electrode connected to the negative teminal of a battery in a discharge tube is called _______. |
| Answer» Correct Answer - Cathode | |
| 2230. |
Electronic configuration of `._23 V^(3+)` ion is `[Ar]^18 3 d^2` and not `[Ar] 18 3d^0 4 s^2`. `V^(3+)`ion is diamagnetic in nature.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
|
Answer» Correct Answer - C ( c) `._23 V : [Ar] 3d_3, 4 s_2 , V^(3+), " [Ar] 3d^2`. Also it is paramegnetic due to the presence of two unpaired electrons. |
|
| 2231. |
The fundamental particle present in anode rays produced by `""_(1)H^(1)` is ______. |
| Answer» Correct Answer - Proton | |
| 2232. |
Cathode rays areA. William CrookesB. G . J. StoneyC. R.A. MillikanD. J.J. Thomson |
|
Answer» Correct Answer - D J.J. Thomson discovered cathode rays and Stoney named it electron. |
|
| 2233. |
Calculate no. of photons emitted by `60W` bulb in `10` hours that produces light of wavelength `6.626 Å` |
| Answer» `E = nhv = nhc//lambda implies 60xx10xx60xx60 = nxx(6.626xx10^(-34)xx3xx10^(8))/(6626xx10^(-10))implies n=72xx10^(23)` | |
| 2234. |
The nature of anode rays depends uponA. Nature of electrodeB. Nature of residual gasC. Nature of discharge tubeD. All the above |
|
Answer» Correct Answer - B The nature of anode rays depends upon the nature of residual gas. |
|
| 2235. |
Which of the following statements about the anode rays is (are) wrong ?A. The anode rays originate from the cathode .B. The anode rays are deflected towards the negative plate of an electric field.C. The e/m ratio for the particles of the anode rays is the same for all gases used in the discharge tube .D. The anode rays can rotate a light paddle wheel placed in their path. |
| Answer» Correct Answer - C | |
| 2236. |
Number of sub-shell in a shell is equal to the number of shel. According to Summerfield : `(n)/(f) = ("Length of major axis")/("Length of minor axis")`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
|
Answer» Correct Answer - A (a) `k = 1` to `n` and cannot be zero :. For fourth shell `k = 1, 2,3, 4,` i.e., four sub-shells. |
|
| 2237. |
`0.24eV` of energy is required to stop photoelectron emitted by radiation of wavelength `2480 Å`. Find work function of metal used? |
|
Answer» Let `phi eV` =work funcion `K.E =0.24 eV =(1240)/(248)-phiimpliesphi=4.76 eV` |
|
| 2238. |
Which of the following graphs correctly represents the variation of particle momentum with associated de Broglie wavelength?A. B. C. D. |
|
Answer» Correct Answer - D `lambda = (h)/(p) rArr lambda. P =` constant graph |
|
| 2239. |
The kinetic energy of an electron accelerated from rest through a potential difference of 5V will beA. `5J`B. `5erg`C. `5eV`D. `8 xx 10^(-10)eV` |
|
Answer» Correct Answer - C `(1)/(2)mv^(2) = ev, V^(2) = (2eV)/(m), V = sqrt((2eV)/(m))` |
|
| 2240. |
what will be the wavelength of a ball of mass 0.1 kg moving with a velocity of `10 m s^(-1)` ? |
|
Answer» According to de-Broglie equation `lambda = (h)/(mv) = (6.626 xx 10^(-34)Js)/((0.1kg)(10ms^(-1))) = 6.626 xx 10^(-34) m(J = kgm^(2)s^(-2))` |
|
| 2241. |
The uncertainity in the positions of an electron and proton is equal, the ratio of the uncertainities in the velocity of an electron and proton isA. `10^(3):1`B. `1:1836`C. `3672:1`D. `1836:1` |
|
Answer» Correct Answer - D `(Deltax_(e).m_(e).Deltav_(e^(-)))/(Deltax_(p).m_(p).Deltavp) =1, (Deltav_(e))/(Deltavp) = (m_(p))/(m_(e))` |
|
| 2242. |
Which properties of electromagnetic radiation are inversely related?A. Amplitude and frequencyB. Energy and wavelengthC. Energy and frequencyD. Wavelength and amplitude |
|
Answer» Correct Answer - B |
|
| 2243. |
An electron is located with an uncertainity equal to its uncertainity in velocity (symbols have usual meaning)?A. `sqrt((h)/(4pi))`B. `(h)/(4pi)`C. `(1)/(2m) sqrt((h)/(pi))`D. Not possible |
|
Answer» Correct Answer - D |
|
| 2244. |
`1 mol` of photons each of frequency `250s^(-1)` would have approximately a total enegry of |
|
Answer» Correct Answer - 1 Total energy `=N_(0)hv` |
|
| 2245. |
energy of one mol of photons whose frequency is `5xx10^(14)Hz` is approximately equal to : |
|
Answer» Energy (E) of one photons `E = hv` `h = 6.626 xx 10^(-34) Js` `v = 5 xx 10^(14)s^(-1)` (given) `E (6.626 xx 10^(-34)Js) xx (5 xx 10^(14)s^(-1))` `= 3.313 xx 10^(-19)J` Energy of one moles of photons `=(3.313 xx 10^(-19)J) xx (6.022 xx 10^(23)mol^(-1))` `=199.51 kJ mol^(-1)` |
|
| 2246. |
The ionization energy of H atom is x kJ. The energy required for the electron to jump from `n = 2` to `n = 3` will be:A. `5x`B. `36x//5`C. `5x//36`D. `9x//4` |
|
Answer» Correct Answer - C `E_("ionisation") =- E_(1)` `E_(1) =- x, E_(2) =- (x)/(4), E_(3) =- (x)/(9)` `DeltaE = (-x)/(9) -(x)/(4)=(x)/(4)-(x)/(9) = (5x)/(36)` |
|
| 2247. |
A 100 watt buble emits monochromatic light of wavelength 400 nm. Then the number of photons emitted per seccond by the buble is nearly - |
|
Answer» Power of the bulb = 100 watt `= 100 Js^(-1)` Energy of one photon `= 4.969 xx 10^(-19)J` Number of photons emitted `=(100 Js^(-1))/(4.969 xx 10^(-19)J) = 2.012 xx 10^(20) s^(-1)` |
|
| 2248. |
For three different metals A,B,C photoemission is observed one by one. The graph of maximum kinetic energy versus frequency f incident radiation are sketched as:A. B. C. D. |
| Answer» Correct Answer - D | |
| 2249. |
Which describes orbital :-A. `Psi`B. `Psi ^(2)`C. `|Psi^(2)|Psi`D. All |
|
Answer» Correct Answer - B `lambda_(m)=lambda_(e)" "lambda=(h)/(mv)` `(h)/(m_(c)v_(c))=(h)/(m_(n)-v_(n))" "(v_(e))/(v_(n))=(m_(n))/(m_(c))` |
|
| 2250. |
The shotest wave length in H spectrum of Lymen series when `R_(H)`=`109678cm^(-1)` is.A. `1002.7 Å`B. `1215.67 Å`C. `1127.30 Å`D. `911.7 Å` |
|
Answer» Correct Answer - D For Lymen series `n_(1)=1` For shortest wave length of Lymen series the energy differnece in two levels showing transition should be maximum `(i.e.n_(2)=oo)" "(1)/(lambda)=R_(H)[(1)/(1^(2))-(1)/(oo^(2)]]` `=109678" " =911.7xx10^(-8)" "=911.7Å` |
|