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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.
| 1301. |
Consider the ground state `Cr` atom `(Z = 24)` The number of electron with the azimuthal number `l = 1` and `2` respectively areA. `16 and 5`B. `12 and 5`C. `16 and 5`D. `12 and 4` |
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Answer» Correct Answer - B `Cr1s^(2)2s^(2)2p^(6)3s6(2)3p^(6)4s^(2)3d^(5) (Z = 2 4 for Cr)` `l = 1 rArr`p orbital rArr 12 electron `l = 2 rArr` d orbital rArr 5 electron |
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| 1302. |
What should be the momentum (in gram cm per second) of a particle if its De Broglie wavelength is `1Å` and the value of `h` is `6.6252xx10^(-27)` erg second?A. `6.6252xx10^(-19)`B. `6.6252xx10^(-21)`C. `6.6252xx10^(-24)`D. `6.6252xx10^(-27)` |
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Answer» Given that `lambda = 1 Å =1 xx10^(-8)cm," "h=6.6252xx10^(-27)` erg second `"or" " "p=(6.6252xx10^(-27))/(1xx10^(-8))=6.6252xx10^(-19) "gram" cm//sec`. |
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| 1303. |
The atoms of two element A and B contain the same number of protons but different number of neutrons in their nuclei. What kind of element are A and B? |
| Answer» A and B are isotopes | |
| 1304. |
The corrent schrodger wave equation for an electron with `E ` as total energy and V as potential energy isA. `(del^2 Psi)/(del x^2) + (del^2 Psi)/(del y^2) +(del^2 Psi)/(del z^2) + (8 pi^2)/(mh^2) (E - V) Psi = 0`B. `(del^2 Psi)/(del x^2) + (del^2 Psi)/(del y^2) +(del^2 Psi)/(del z^2) + (8 pi m)/(h^2) (E - V) Psi = 0`C. `(del^2 Psi)/(del x^2) + (del^2 Psi)/(del y^2) +(del^2 Psi)/(del z^2) + (8 pi^2 m)/(h^2) (E - V) Psi = 0`D. `(del^2 Psi)/(del x^2) + (del^2 Psi)/(del y^2) +(del^2 Psi)/(del z^2) + (8 pi^2 m)/(h) (E - V) Psi = 0` |
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Answer» Correct Answer - C ( c) |
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| 1305. |
A sample of hydrogen contains equal number of `H^(1),H^(2) "and"H^(3)` atoms. The ratio of total number of protons and neutrons `((p)/(n))` in the sample is : |
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Answer» Correct Answer - 1 |
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| 1306. |
Ultraviolet light of `6.2eV` falls on Caesium surface (work function `= 1.2eV)`. The kinetic energy (in electron volts) of the fastest electron emitted is approximatelyA. `5eV`B. `4eV`C. `3eV`D. `2eV` |
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Answer» Correct Answer - A `K.E = E-W` |
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| 1307. |
The mass of an electron is `9.1xx10^(-31)kg`. If its K.E. is `3.0xx10^(-25) J`, calculate its wavelength |
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Answer» `K.E = 1/ 2 "mu"^2 ...(1)` ` :. U = h/( n lambda) ...(2) ` By eqs , (1) and (2) , ` K.E. = 1 /2 m h^2 /( m^2 lambda ^2 ) or lambda = sqrt ( h^2 /( 2 m xx K.E.))` ` :. lambda = sqrt (( 6.626 xx 10^(34) xx 6.626 xx 10^(-34))/( 2xx 9.1 xx 10^(-31) xx 3. 0 xx 10^(-25)))` ` = 8966 xx 10^(-10) m = 8966 Å `. |
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| 1308. |
Which of the following should be the wavelength of an electron if its mass is `9.1xx10^(-31)kg` and its velocity is `1//10` of that of light and the value of `h` is `6.6252xx10^(-24)` joule second?A. `2.446xx10^(-7)` metreB. `2.246xx10^(-9)` metreC. `2.246xx10^(-11)` metreD. `2.246xx10^(-13)` metre |
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Answer» `m=9.1xx10^(-31)kg, " "c=(1)/(10)` of velocity of light `=(1)/(10)xx3xx10^(8)` metre/second i.e. `3xx10^(7)` metre/second `h=6.6252xx10^(-34)` joule second `lambda=(h)/(mc)=(6.6252xx10^(-34))/(9.1xx10^(-31)xx3xx10^(7))` `=(6.6252xx10^(-34))/(27.3xx10^(-24))` `=2.426xx10^(-11)` metre |
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| 1309. |
Which experimental evidence most clearly supports the suggestion that electrons have wave properties ?A. DiffractionB. Emission spectraC. Photoelectric effectD. Deflection of cathode rays by a magnet |
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Answer» Correct Answer - A |
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| 1310. |
Select the incorrect graph for velocity of `e^(-)` in an orbit vs. Z, `1/n` and n :A. B. C. D. |
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Answer» Correct Answer - D `v prop (z)/(n)` |
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| 1311. |
What is the frequency of photon, whose momentum is `1.1xx10^(-23) kg ms^(-2)`A. `5xx10^16` HzB. `5xx10^17` HzC. `0.5xx10^18` HzD. `5xx10^18` Hz |
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Answer» Correct Answer - D `because E=mc^2, E=hv` `therefore mc^2 =hv` `v=(mc^2)/(h)=(1.1xx10^(-23)xx3xx10^8)/(6.6xx10^(-34))=5xx10^18` Hz |
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| 1312. |
If radiation corresponding to first line of 'Balmer series' of `He^(+)` ion is subjected to a smaple of `Ki^(+2)` ion (containing atoms in different energy states) and it causes ejection of photoelectron with non-zero kinetic energy then calculate least shell number in which the electron must be present in `Li^(+2)`. |
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Answer» Correct Answer - 5 |
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| 1313. |
Visible light photons do not show Compton effect because theyA. Move very slowlyB. Have no momentumC. Have very less massD. Have larger wavelength |
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Answer» Correct Answer - D `lambda = (h)/(mv)` |
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| 1314. |
The electronic configuration of `H^-` isA. `1s^0`B. `1s^1`C. `1s^2`D. `1s^1 2s^1` |
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Answer» Correct Answer - C Electronic configuration of `H^-` is `1s^2`. It has 2 electrons in extra nuclear space |
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| 1315. |
As the frequency of the light increases, the momentum of its PhotonA. IncreasesB. DecreasesC. Remains sameD. Cannot be predicted |
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Answer» Correct Answer - A `lambda = (h)/(mc)` |
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| 1316. |
A hydrogen atom in the ground state is excited by monochromatic radiation of wavelength `lambda A`. The resulting spectrum consits of maximum 15 different lines. What is the wavelength `lambda`?`A. `937.3 Å`B. `1025 Å`C. `1236 Å`D. None of these |
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Answer» Correct Answer - 1 Total number of spectral lines given by `(1)/(2)[n-1]xxn=15," ":.n=6` Thus, electron is excited upto `6^(th)` energy level from ground state. Therefore, `(1)/(lambda)=R_(H)[(1)/(1^(2))-(1)/(n^(2))]=109737xx(35)/(36),` `lambda=9.373xx10^(-5)cm=937.3Å` |
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| 1317. |
In an atom, the total number of electrons having quantum numbers `n = 4,|m_(l)| = 1` and `m_(s) =- (1)/(2)` is |
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Answer» Correct Answer - 6 n=4 l=0,1,2,3 `|m_l|=1 rArr pm1` `m_s=-1/2` For l=0, `m_l=0` `l=1 , m_l =-1,0,+1` `l=2 , m_l=-2,-1,0,+1,+2` `l=3, m_l=-3,-2,-1,0,+1,+2,+3` So, six electrons can have `|m_l|` =1 and `m_s=-1/2` |
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| 1318. |
The hydrogen atom in the ground state is excited by mass of monochromatic radiations of wavelength ` lambda Å ` . The resulting spectrum consists of maximum `15` different lines . What is the value of ` lambda` ? (`R_H = 109737 cm^(-1))`.A. `937.3 overset(0)(A)`B. `1025 overset(0)(A)`C. `1236 overset(0)(A)`D. `618 overset(0)(A)` |
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Answer» Correct Answer - A Transition is `6 rarr 1, (1)/(lambda) = R_(H)z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` |
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| 1319. |
The emission spectrum of hydrogen in the visible consists of :A. a continuous band of light .B. a series of equally spaced lines .C. a series of lines that are closer at low energies .D. a series of lines that are closer at high energies . |
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Answer» Correct Answer - D |
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| 1320. |
The electronic configuration of silver atom in ground state is.A. `[Kr]3d^10 4s^1`B. `[Xe]4 f^14 5d^10 6s^1`C. `[Kr]4d^10 5s^1`D. `[Kr]4d^9 5s^2` |
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Answer» Correct Answer - C The electronic configuration of Ag in ground state is `[Kr] 4d^10 5s^1` |
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| 1321. |
Hydrogen spectrum consists of ……….. |
| Answer» Correct Answer - Five series of line | |
| 1322. |
The mangnitue of spin angular momentum of electron is givenby :A. ` s= sqrt ( s(s+1) )1/( 2pi)`B. ` s=s h/( 2 pi)`C. `s=+- 1/2 . h/(2pi)`D. `s= sqrt((3)/(2) . (h)/(9) 2pi)` |
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Answer» Correct Answer - A::D Spong angualr momentum `= sqrt (S(s+ 1)). h/(2pi)` ` s= sqrt (1/2 (1/2 + 1)) . h/(2pi)` ` = (sqrt 3)/2 . h/(2pi)`. |
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| 1323. |
Which of the following statements is not correct for an electron that has the quantum numbers n=2 and m=2A. The electron may have the quantum number `s=+1/2`B. The electron may have the quantum number `l=2`C. The electron may have the quantum number `l=3`D. The electron may have the quantum number `l=0,1,2,3` |
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Answer» Correct Answer - D m cannot be greater than l (=0,1) |
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| 1324. |
The coloured radition with lowest energy isA. RedB. BlueC. GreenD. Yellow |
| Answer» Correct Answer - Red | |
| 1325. |
Which of the following represents the correct sets of the four quantum numbers of a 4d electronA. `4,3,2,1/2`B. `4,2,1,0`C. `4,3,-2,+1/2`D. `4,2,1,-1/2` |
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Answer» Correct Answer - D For 4d, n=4, l=2, m=-2, -1, 0,+1,+2 , `s=+1/2` |
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| 1326. |
The mangnitue of spin angular momentum of electron is givenby :A. `S = sqrt(s(s + 1)) (h)/(2 pi)`B. `S = (h)/(2 pi)`C. `S = (sqrt(3))/(2) xx (h)/(2 pi)`D. `S = +- (1)/(2) xx (h)/(2 pi)` |
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Answer» Correct Answer - C ( c) Spin angular momentum `= sqrt(s(s +1)) (h)/(2 pi)` `S = sqrt((1)/(2)((1)/(2) +1)) (h)/(2pi) = (sqrt(3))/(2) xx (h)/(2 pi)`. |
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| 1327. |
What is the orbit angular momentum of a `d` electron?A. `(6 h)/(2 pi)`B. `(sqrt(6) h)/(2 pi)`C. `(12 h)/(2 pi)`D. `(sqrt(12) h)/(2 pi)` |
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Answer» Correct Answer - B (b) We know that for `d-"electron" l = 2` `mu = sqrt(l(l + 1)) (h)/(2 pi)` , `mu = sqrt(2(2 + 1)) (h)/(2 pi)` `mu = sqrt(2(2 + 1)) (h)/(2 pi)` , `mu = sqrt(6) (h)/(2 pi)`. |
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| 1328. |
Calculate the wavelength of an ubraviolet wave , if its frequency is `12 xx 10^(16)` cycles per second and `c = 3 xx 10^(6) "in" s^(-1)`? |
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Answer» Velocity of radition `= 3 xx 10^(8)m s^(-1)` Frequency of radition `= 12 xx 10^(16) Hz` Wavelength `(lambda) = (c )/(v) = (3 xx 10^(8))/(12 xx 10^(16)) = 0.25 xx 10^(8-) m` ` or lambda = 2.5 xx 10^(-8)m` |
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| 1329. |
A photon in X region is move emergetic than in the visible region X isA. `18` raysB. `UV` raysC. MicrowavesD. Rediowaves |
| Answer» Correct Answer - UV | |
| 1330. |
Find the ratio of frequency of violet light `(lambda_(1) = 4.10 xx %^(-5) cm)` to that of red light `(lambda_(2) = 6.56 xx 10^(-5) cm)` Also determine the ratio of energies carried by them |
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Answer» Using `c= v lambda` where c is speed of light v is frequency `lambda` is wavelength `(v_(1))/(v_(2)) = lambda_(2)/(lambda_(1)) [1 : "violet and 2 : red"]` `rArr (v_(1))/(v_(2)) = (6.56 xx 10^(-5))/(4.10 xx 10^(-5)) = 1.6: 1` Now the energy assciated with electromagnitic radiation is given by `E = hv` .THerefore `(E_(1))/(E_(2)) = (v_(1))/(v_(2)) = (lambda_(1))/(lambda_(2)) = 1.6 : 1` Hence the ratio of energies is same as that of frequencies. |
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| 1331. |
The maximum number of electrons that can be accommodated in an orbital isA. oneB. twoC. threeD. four |
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Answer» Correct Answer - B (b) Each orbital has at least two electrons. |
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| 1332. |
Which of the following sets of quantum number is allowableA. `n = 2,l = 1,m = 0,s = +1//2`B. `n = 2,l = 1,m = -1,s = -1//2`C. `n = 2,l = -2,m = 1,s = +1//2`D. `n = 2,l = 1,m = 0,s = 0` |
| Answer» Correct Answer - A | |
| 1333. |
The electronic configuration of silver atom in ground state is.A. `[Kr] 3d^10 4 s^1`B. `[Xe] 4f^14 5 d^10 6 s^1`C. `[Kr] 4 d^10 5 s^1`D. `[Kr]4 d^9 5 s^2` |
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Answer» Correct Answer - C ( c) The electronic configuration of `Ag` in ground state is `[Kr] 4d^10 5 s^1`. |
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| 1334. |
The two electrons in K sub-shell will differ inA. Principal quantum numberB. Azimuthal quantum numberC. Magnetic quantum numberD. Spin quantum number |
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Answer» Correct Answer - D The two electrons will have opposite spins |
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| 1335. |
The maximum number of electron inm`n = 1,l = 0, m = 0,s= +- 1//2 ` is ………. |
| Answer» Correct Answer - Two | |
| 1336. |
The orbital having `n = 2,l=p 12` and `n= 0`will be desigrated asA. `2p_(z)`B. `2p_(x)`C. `2p_(y)`D. `3d_(z)2` |
| Answer» In the second or I energy level `(n= 2),1= 1` for p orbit `m= 90`x axis hence for orbit will be designated as `2p_(2)` | |
| 1337. |
The Vividh Bharati station broadcasts on a frquency of `1368kHz`. Calculate the wavelength of electromag-netic radiation emitted by the transmitter. |
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Answer» `lambda = (c )/(v)` `c=3xx10^(8)ms^(-1),v=1368kHz=1368xx10^(3)Hz("or"s^(-1))` `therefore " "lambda=((3xx10^(8)ms^(-1)))/((1368xx10^(3)s^(-1)))=219.3m` |
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| 1338. |
Maximum number of electrons in a sub-shell with `l = 3` and `n = 4` is.A. 14B. 16C. 10D. 12 |
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Answer» Correct Answer - A (a) n represents the main energy level and l represents the subshell. If n=4 and l=3, the subshell is 4f. In f-subshell, there are 7 orbitals and each orbital can accommodate a maximum number of two electron, so maximum number of electrons in 4f subshell `=7xx2=14`. |
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| 1339. |
Which of the following is associated will the orbital desigrated by `n = 2,l = 1 `?A. SphericalB. TetrahedralC. Dumb-shellD. Pyramidil |
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Answer» Correct Answer - C `l = 1` therefore p orbitals |
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| 1340. |
A base ball of mass `200 g` is moving with velocity of `3 xx 10^(3) cm s^(-1)` .If we can locte the base ball with an error equal to the magnitude of the wavelength of the light used `(5000 Å)` how wil the uncertainty in momentum be used with the total momentum of the base ball? |
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Answer» Momentum of the basic ball = mv `p = 200 xx 10 = 6 xx 10^(5) cm g s^(-1)` `Delta p Delta x ge h//4pi` `Delta p ge (h)/(4pi Delta x) ( Deltax = 5000 xx 10^(-8) = 5 xx 10^(-5))` ` (DeltaP)/Pge (6.626 xx 10^(-27))/(4 xx 3.14 xx 5 xx 10^(-5))` `( Delta p)/(p) ge (6.626 xx 10^(-27))/(4 xx 3.14 xx 5 xx 10^(-5))` `= 1.756 xx 10^(-29)` |
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| 1341. |
How many electrons can fit into the orbitals that comprise the `3^(rd)` quantum shell `n = 3` ?A. 2B. 8C. 18D. 32 |
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Answer» Correct Answer - C ( c) `n = 3` electrons in `3^( rd) shell = 2 n^2` =`2 xx 32 = 18`. |
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| 1342. |
What is the maximum number of electrons that can be associated with a following set of quantum numbers ? `(n = 3, l = 1 and m = -1)`.A. 10B. 6C. 4D. 2 |
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Answer» Correct Answer - D (D) The orbital of the electron having n=3, l=1 and m=-1 is `3p_(z)` (as `nl_(m)`) and an orbital can have a maximum number of two electrons with opposite spins. `:. 3p_(z)` orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1. |
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| 1343. |
If a light of wavelght `lambda` hits moving electron the uncertainty in measurement of its position will beA. Greater than `lambda`B. Less than `lambda`C. Equal to `lambda `D. Any value |
| Answer» Correct Answer - Equal to `lambda ` | |
| 1344. |
The number of orbitals is a quantum shell is equal to …………….. |
| Answer» Correct Answer - `m=n^(2)` | |
| 1345. |
The dual nature of radiation was proposed by ……….. |
| Answer» Correct Answer - de Broglie | |
| 1346. |
A sample of hydrogen gas has same atom in out excited state and same atom in other excited state it emits three difference photon.When the sample was irradiated with radiation of energy `2.85 eV` ,it emits `10` different photon all having energy in or less than `13.6 eV` ltbrtgt a. Find the principal quantum number of initially excited electrons b. Find the maximum and minimum energies of the initially emitted photon |
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Answer» Initailly the sample emit three difference photon it means for one of the ecxited states `n = 3` For the other excited state `n = 1 or n = 2` After transition it emit `10` diofference photon it means for the excited state `n = 5` Energy of fifth energy level `= (-13.6)/((5)^(2)) = - 0.544 eV` Energy of total energy level `= (-13.6)/((3)^(2)) = - 1.51 eV` Energy of second energy level `= (-13.6)/((2)^(2)) = - 3.4 eV` So, after absurption of `2.85 eV` energy only electron from accond energy level can jump to fifth energy level principal quantum number of initially excited state are `2` and `3` respectively Maximum energy of emitted photon `= 13.6 - 1.51 = 12.09 eV` Maximum energy of emitted photon `= 3.4 - 1.51 = 1.89 eV` |
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| 1347. |
The transitionis `He^(o+)` ion that would have the same wavelength as the first Lyman line in hydtrogen spectrum isA. `2 rarr 1`B. `5 rarr 3`C. `4 rarr 2`D. `6 rarr 4` |
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Answer» Correct Answer - C `bar v _(He^(o+)) = (1)/(lambda) = R xx 2^(2) ((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))` `= R ((4)/(n_(1)^(2)) - (4)/(n_(2)^(2)))`…..(i) `bar v _(theta_(2)^(o+)) = (1)/(lambda) = R xx ((1)/(1^(2))- (1)/(2^(2)))`….(ii) Compare equation (i) and (ii) we get `(1)/(1^(2)) = (4)/(n_(1)^(2)), n_(1)^(2)= 4 n_(1) = 2` `(4)/(n_(2)^(2)) = (1)/(2^(2)),n_(2)^(2)= 16 n_(2) = 4` `( h rarr 2)` |
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| 1348. |
A hydrogen like atom in ground st6ate abserbs n photon having the same energy and its emit exacity n photon when electron transition tekes placed .Then the energy of the absorbed photon may beA. `91.8 eV`B. `40.8 eV`C. `48.4 eV`D. `54.4 eV` |
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Answer» Correct Answer - A::B Exp: Number of dark line (inabsorption ), i.e. excittation = Number of bright lines its emission i.e. de -excitation It is possible only when the e is excitated `n = 2` from ground state Clearly `Delta E = 91.8 eV ` and `40.8 eV` are possible `Li^(2+) (He^(o+))` |
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| 1349. |
Calculate the uncertainty in the position `(Delta x)` of an electron if `Delta v is 0.1 %` .Take the velocity of electron `= 2.2 xx10^(6) ms^(-1)` and mass of electron as `9.108 xx 10^(-31)kg` |
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Answer» Gives `Delta v = 0.1 % `of the velocity of the electron `= (0.1)/(100) xx 2.2 xx 10^(6) m s^(-1)` `Delta x xx m Delta v = (h)/(4pi)` or `Delta x = (6.63 xx 10^(-34) Js)/( 4 xx 3.14 xx 9.108 xx 10^(-31) kg xx 2.2 xx 10^(2) ms^(-1))` `= 0.02624765 xx 10^%(-6) m` `= 262.4765 xx 10^(-10) m` Since `Delta x ` is much longer than the atomic diameter `(+10^(-10)m`1 the uncertainty principle is applicable in this case. |
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| 1350. |
A hydrogen like atom in ground state absorbs `n` photon having the same energy and its emits exacity `n`` photon when electrons transition takes placed .Then the energy of the absorbed photon may beA. 91.8 eVB. 40.8 eVC. 48.4eVD. 54.4 eV |
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Answer» Correct Answer - a b |
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