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.
| 2401. |
Statement-1: The angular momentum of `e^(-) " in " 4 f " orbital is " sqrt3(h)/(pi)`. Statement-2: Augular momentum of electron in 4 th orbit is `(2h)/(pi)`. |
|
Answer» Correct Answer - b |
|
| 2402. |
Calculate the minimum kinetic energy in eV of photoelectron produced in caesium by 400 nm light . The critical (maximum) wavelength for the photoelectric effect in caesium is 660 nm, when the potential difference is 1.78 V. |
|
Answer» Correct Answer - 3 |
|
| 2403. |
Statement -I : When projected to words towards as atoms having heavy nucleus , most of the `alpha` particle will pass through it without appereciable decviation. ltbr. Statement -II : All the positive charge of atom is located in a small dense located at center.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 - A | |
| 2404. |
Statement -I : The angular momentum of an electron in `n^(th)` orbital it same for all H-like species . Statement -II : The velocity of electron in `m^(th)` orbit is equal for all H-like species.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 | |
| 2405. |
Electronmagnetic radiations having `lambda=310Å`are subjected to a metal sheet having work function `=12.8eV`. What will be the velocity of photoelectrons with maximum Kinetic Energy….A. 0, no emission will occurB. `4.352 xx 10^(6)m//s`C. `3.09 xx 10^(6)m//s`D. `8.72 xx 10^(6)m//s` |
|
Answer» Correct Answer - C `(hc)/(lambda) = phi +(1)/(2)mv^(2)` |
|
| 2406. |
A 1-kW radio transmitter operates at a frequency of 880 Hz. How many photons per second does it emit?A. `1.71 xx 10^(21)`B. `1.71 xx 10^(30)`C. `6.02 xx 10^(23)`D. `1.71 xx 10^(33)` |
|
Answer» Correct Answer - D `E = n. (hc)/(lambda) (1kW = 1000 J//s)` |
|
| 2407. |
If `10^(-17)J` of energy from mnochromatic light is needed for exciting electro magnetic oscilators. If one electro magnetic oscilator absorbs one photon how many electro magnetic oscilator can be excited?A. 27B. 28C. 29D. 30 |
|
Answer» Correct Answer - B `E = n.(hc)/(lambda)` |
|
| 2408. |
Which of the following statement (s) is `//`are true?A. If H-atoms are in 3rd energy level then number of maximum different spectral lines produced by 2 atoms will be equal to number of maximum spectral lines produced by 3 H-atoms.B. For third excited state, `He^(+)` ions and maximum number of different spectral line produced by 4 H-atoms are equal.C. Energy evolved from 3rd to 2nd energy level transition in `He^(+)` can be used ot ionise `Li^(2+)` from ground level.D. Energy evolved by a `He^(+)` ion which is in its 1st excited state can ionise a hydrogen atom which is in its ground state. |
|
Answer» Correct Answer - a,b,d |
|
| 2409. |
Aufbau principle is obeyed in which of the following electronic configurationA. `1s^2 2s^2 2p^6`B. `1s^2 3p^3 3s^2`C. `1s^2 3s^2 3p^6`D. `1s^2 2s^2 3s^2` |
|
Answer» Correct Answer - A According to Aufbau principle electron are filling increasing order of energy. Therefore the electronic configuration `1s^2 2s^2 2p^6` obeys Aufbau principle |
|
| 2410. |
`S_(1)` : Bohr model is applicable for `Be^(2+)` ion . `S_(2)` : Total energy coming out of any light source is integral multiple of energy of one photon. `S_(3)` : Number of waves present in unit length if wave number . `S_(4)` : e/m ratio in cathode ray experiment is independent of the nature of the gas . Select the correct set of True-False for above statement.A. FFTTB. TTFFC. FTTTD. TFFF |
|
Answer» Correct Answer - C |
|
| 2411. |
Bohr theory is applicable toA. `He`B. `Li^(2+)`C. `He^(2+)`D. `H-`atom |
| Answer» Correct Answer - A::C | |
| 2412. |
Iso-electronic species areA. `F^- , O^(-2)`B. `F^- , O`C. `F^- , O^+`D. `F^- , O^(+2)` |
| Answer» Correct Answer - A | |
| 2413. |
Which is not isoelectronic with the other threeA. COB. `NO^+`C. `CN^-`D. `O_2` |
| Answer» Correct Answer - D | |
| 2414. |
Wave length associated with electron motionA. Increase with increase in speed in electronB. Remains same irrespective of speed of electronC. Decreases with increase in speed of `e^-`D. Is zero |
| Answer» Correct Answer - C | |
| 2415. |
The wavelengths of electron waves in two orbits is 3:5. The ratio of kinetic energy of electrons will beA. `25:9`B. `5:3`C. `9:25`D. `3:5` |
|
Answer» Correct Answer - A `(lambda_(1))/(lambda_(2)) = sqrt((KE_(2))/(KE_(1)))` |
|
| 2416. |
An element has 8 electrons in the valence shellA. it will lose electronB. it will gain an electronC. It neither gains or lose electronD. it will make bond with itself |
| Answer» Correct Answer - C | |
| 2417. |
An element has 2 electrons in K shell, 8 electrons in L shell, 13 electrons in M shell and one electron in N shell. The element isA. `Cr`B. `Fe`C. `V`D. `Ti` |
|
Answer» Correct Answer - A Magnetic moment `=sqrt(n(n+2))BM` |
|
| 2418. |
One quantum is absorbed per molecule of gaseous iodine for converting into iodine atoms. If light absorbed has wavelength of `5000A^(@)`, The energy required in `kJ mol^(-1)` isA. 139B. 239C. `23.9`D. 60 |
|
Answer» Correct Answer - B `E = N_(0)hv = (N_(0)hc)/(lambda)` `=(6.02 xx 10^(23) xx 6.626 xx 10^(-34) xx 3xx 10^(8))/(5000 xx 10^(-10)) = 239.3 kJ mol^(-1)` |
|
| 2419. |
Rutherfords experiments , which established the nuclear model of atom , used a beam of:-A. `beta`-particles, which impinged on a metal foil and got abosorbedB. `gamma`-rays, which impinged on a metal foil and ejected electronsC. Helium atoms, which impinged on a metal foil and got scatteredD. Helium nuclei, which impinged on a metal foil and got scattered |
|
Answer» Correct Answer - D Heavy `alpha` particles relative to `beta` particle can given more information about nuclear parameter. |
|
| 2420. |
The spin magnetic momentum of electron in an ion is `4.84BM`. Its total spin will beA. `+-1`B. `+-2`C. `ge sqrt((h)/(4pi))`D. `+-2.5` |
|
Answer» Correct Answer - B `mu = sqrt(4S(S+1))` |
|
| 2421. |
calculate the mass of a photon with wavelength 3.6 A |
|
Answer» `lambda = 3.6 A^(0) = 3.6 xx 10^(-10)m` Velocity of photon = velocity of light `m = (h)/(lambda v) = (6.626 xx 10^(-34)Js)/((3.6 xx 10^(-10)m)(3 xx 10^(8)ms^(-1)))` `=6.135 xx 10^(-33)kg` |
|
| 2422. |
Arrange the following atomic numbers in the increasing order of number of unpaired electronsA. 16B. 32C. 40D. 57 |
|
Answer» Electronic configuration is ` 1s^2 2s^2 2p^6 3s^2 3p^4` number of unpaired electrons is 2. Electronic configuration is Electronic configuration is ` 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^2`number of unpaired electrons is 2. Electronic configuration is ` 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^2` number of unpaired electrons is 2. Electronic configuration is `[Xe] 6s^2 5d^1` number of unpaired electrons 1 `therefore` order is (i) = (ii)= (iii) gt (iv) |
|
| 2423. |
Total Number of unpaired electrons in `d-` orbitals of an atom element of atomic number 29 is `:`A. 10B. 1C. 0D. 5 |
|
Answer» Correct Answer - C Atomic number of Cu is 29=`(Ar)4s^1 3d^10` |
|
| 2424. |
Explain why the uncertainty principle has significated when applied to macroscope objects such as moving car ? |
|
Answer» Correct Answer - A In case of macroscapic the energy of photons cannot disturb the position and momentum of the object be such case the uncertainqainies on position amd moment are negligble as compared to the size of the object and moment of the object, respectively |
|
| 2425. |
The electronic configuration of an element is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` .This represents itsA. Excited stateB. Ground stateC. Cationic formD. Anionic form |
|
Answer» Correct Answer - B `3d^5 4s^1` system is more stable than `3d^4 4s^2` , hence former is the ground state configuration |
|
| 2426. |
An electron has a spin quantum number `+1(1)/(2)` and a magnetic quantum number -1. It cannot be present inA. d-orbitalB. f-orbitalC. p-orbitalD. s-orbital |
|
Answer» Correct Answer - D (d) Spatial orientation of the orbital with respect to standard set of cordinate axis. Magnetic quantum number -1 is possible only when the azimuthal quantum number have value l=1, which is possible for p, d and f-subshells but not for s-subshell because the value of l for s-subshell is zero. |
|
| 2427. |
If an electron has spin quantum number of `-(1)/(2) ` and magnetic quantum number of -1 it cannot be present in:A. d-orbitalB. f-orbitalC. p-orbitalD. s-orbital |
|
Answer» Correct Answer - D m=-1 is not possible for s orbital (l=0) |
|
| 2428. |
An electron, a proton and an alpha particle have kinetic energy of `16E,4E` and `E` respectively. What is the qualitavtive order of their de Broglie wavelengths :-A. `lambda_(e)gtlambda_(p)=lambda_(alpha)`B. `lambda_(p)=lambda_(alpha)gtlambda_(e)`C. `lambda_(p)gtlambda_(e)gtlambda_(alpha)`D. `lambda_(alpha)ltlambda_(e)gtgt lambda_(alpha)` |
|
Answer» Correct Answer - A `lambda_(e):lambda_(p):lambda_(alpha)=(1)/sqrt(m_(e)1.6E):(1)/(sqrtm_(p),4E):(1)/(sqrtm_(alpha.E))` Hence `lambda_(e):lambda_(p)=lambda_(alpha)` |
|
| 2429. |
An electron, a proton and an alpha particle have kinetic energy of `16E,4E` and `E` respectively. What is the qualitavtive order of their de Broglie wavelengths :-A. `lambda_(e)gtlambda_(p)=lambda_(alpha)`B. `lambda_(p)=lambda_(alpha)gtlambda_(e)`C. `lambda_(p)gtlambda_(e)gtlambda_(alpha)`D. `lambda_(alpha)ltlambda_(e)gtgtlambda_(alpha)` |
|
Answer» Correct Answer - A `Cu^(+)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)` all the electron are paired , hence it is paramagnatic |
|
| 2430. |
An electron, a proton and an alpha particle have kinetic energy of `16E,4E` and `E` respectively. What is the qualitavtive order of their de Broglie wavelengths :-A. `lambda_(e) gt lambda_(p) = lambda_(alpha)`B. `lambda_(p) = lambda_(alpha) = alpha_(e)`C. `lambda_(p) gt lambda_(e) gt lambda_(alpha)`D. `lambda_(e) gt lambda_(alpha) gt lambda_(p)` |
|
Answer» Correct Answer - A |
|
| 2431. |
From the following sets quantum number state which are possible. Explain why the other are not permitted ? a. `n = 0, l = 0, m= 0, s = + 1//2` b. `n = 1, l = 0, m= 0, s = - 1//2` c. `n = 1, l = 1, m= 0, s = + 1//2` d. `n = 1, l = 0, m= +1, s = + 1//2` e. `n = 0, l = 1, m= -1, s = - 1//2` f. `n = 2, l = 2, m= 0, s = - 1//2` g. `n = 2, l = 1, m= 0, s = - 1//2` |
|
Answer» a. is not posible because `n = 0` is wrong b.is posible c. is not posible because `l = n` is wrong d. is not posible because `m = + 1` is wrong e. is not posible is not posible because f. is not posible because `l = 2` is wrong is posible. |
|
| 2432. |
The ratio of wave length of photon corresponding corresponding to the `alpha`-line of Lyman series in H-like and beta-line of Balmer series in `He^(+)` isA. `1:1`B. `1:2`C. `1:4`D. `3:16` |
|
Answer» Correct Answer - A `(lambda_(alpha))/(lambda_(beta))`=`(R.1^(2)((1)/1^(2)-(1)/(2^(2))))/(R.2^(2)((1)/(2^(2))-(1)/(4^(2))))=1` |
|
| 2433. |
Which of the following sets of quantum number is//are permitted?A. `n= 3,l= 3,m= 0,s= (1)/(2)`B. `n= 3,l= 2,m= 2,s= -(1)/(2)`C. `n= 3,l= 1,m= 2,s= -(1)/(2)`D. `n= 3,l= 0,m= 0,s= +(1)/(2)` |
| Answer» When `n= 3,l` cannot be `3` so a is not permitted when `l = 1 ,m` cannot be `= +2` so c is not permitted Hence a and c are corerect | |
| 2434. |
In hydrogen spectrum the different lines of Lyman series are present isA. UV fieldB. IR fieldC. Visible fieldD. Far IR field |
|
Answer» Correct Answer - A According to Hydrogen spectrum series |
|
| 2435. |
The ratio of the frequency corresponding to the third line in the lyman series of hydrogen atomic spectrum to that of the first line in Balmer series of `Li^(2+)` spectrum isA. `4/5`B. `5/4`C. `4/3`D. `3/4` |
|
Answer» Correct Answer - D Lyman series , `n_1=1` For third line of Lyman series , `n_2=4` For hydrogen , Z=1 `upsilon_H=c/lambda=c. R_H Z^2 (1/n_1^2 -1/n_2^2)` `=c.R_H (1)^2 (1/1 -1/(4)^2) =15/16 R_H .c` For lithium , Z =3 For first line of balmer series , `n_1=2 , n_2 =3` `upsilon_(Li^(2+)) =c.R_H(1)^2 (1/1-1/(4)^2)=15/16R_H . c` For lithium, Z=3 For First line of balmer series , `n_1=2, n_2=3` `upsilon_(Li^(2+))=c. R_H (3)^2 (1/(2)^2 -1/(3)^2)=c. R_Hxx9xx5/36` `=5/4c. R_H` `upsilon_H/upsilon_(Li^(2+)) =((15//16)cR_H)/((5//4)cR_H)=15/16xx4/5=3/4` |
|
| 2436. |
A boll of mass `200 g` is moving with a velocity of ` 10 m sec^(-1)` . If the error in measurement of velocity is ` 0. 1%`, the uncertainty in its position is :A. ` 3. 3 xx 10^(31) m`B. ` 3.3 xx 10^(-27) m`C. ` 5.3 xx 10^(-25) m`D. ` 2.64 xx 10^(-32)m` |
|
Answer» Correct Answer - D `Delta u = ( 0.1)/(100 ) xx 10 =10^(-2) m sec ^(-7)` Now ` Delta u. Delta x = h/(4 pim)` ` therefore Delta x = (6.625 xx 10^(-34))/(4 xx 10^(-2) xx 3. 14 xx 200 xx 10^(-3))` ` = 2. 46 xx 10^(-32) m`. |
|
| 2437. |
The Lyman series of the hydrogen spectrum can be represensted by the equation `v = 3.2881 xx 10^(15)s^(-1)[(1)/((1)^(2)) - (1)/((n)^(2))] [where n = 2,3,…….)` Calculate the maximum and minimum wavelength of the lines in this series |
|
Answer» `bar v = (1)/(lambda) = (v)/(c ) = (3.2881 xx 10^(15))/(3 xx 10^(8)) m^(-1)[(1)/((1)^(2)) - (1)/(n^(2))]` Wavelength is maximum `(bar v_(min))` when n is minimum so that `1//n^(2) `is maximum : `bar v_(min) = (1)/(lambda_(max)) = (3.2881 xx 10^(15))/(3 xx 10^(8)) ((1)/(1)^(2) - (1)/((2)^(2)))` `:. lambda_(max) = (3 xx 10^(8))/(3.2881 xx10^(15)) xx (4)/(3)` `= 12165 xx 10^(-7) m = 121.67 nm ` Wavelength is minimum `(bar v_(max))` when n is `oo` i.e.series converges `:. v_(max) = (1)/(lambda_(max)) = (3.2881 xx 10^(15))/(3 xx 10^(8))` `:. lambda_(max) = 0.9124 xx 10^(-2) m = 91.24 nm` |
|
| 2438. |
A stationary hydrogen atom emits a photon corresponding to the first line of the Lyman series. What is the velocity of recoil of the atom? `(m_(H)=1.672xx10^(-27) kg)` |
|
Answer» Correct Answer - 3.25 m/s |
|
| 2439. |
The Lyman series of hydrogen spectrom can be respectively by the equation `v = 3.28 xx 10^(15)[(1)/(1^(2)) - (1)/(n^(2))] s^(-1)` Calculate the maximum and minimum frequencies in this series |
|
Answer» Correct Answer - A::B::C::D Lyman frequrncy will be maximum corresponding to the maximum energy transition i.e. `1 rarr oo` `rArr v_(max) = 3.28 xx 10^(15) [(1)/(1^(2)) - (1)/(oo^(2))] s^(-1) = 3.26 xx 10^(15) s^(-1)` Note that coreresponding wavelength will be shortest wavelength And Lyman frequency will be coreresponding to minimum energy transition i.e. `1 rarr 2` `rArr v_(min) = 3.28 xx 10^(15) [(1)/(1^(2)) - (1)/(2^(2))] s^(-1) = 2.46 xx 10^(15) s^(-1)` Note that coreresponding wavelength will be longest wavelength. |
|