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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.
| 501. |
the threashold frequency `v_(0)` for a metal is `7xx10^(14)s^(-1)` . Calculate the kinetic energy of an electron emitted when radiation of fequency `v = 1.0 xx 10^(15)s^(-1)` hits the metal .A. `1.988 xx 10^(-17)J`B. `1.988 xx10^(19)J`C. `3.988 xx 10^(-19)J`D. `1.988 xx 10^(-19)J` |
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Answer» Correct Answer - D `hv = hv_(0) +kE, K.E = h(v -v_(0))` |
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| 502. |
A strong argument for the particle nature of cathode rays is:A. they can propagate in vacuumB. they produce fluoresceneC. they cast shadoesD. they are deflected by electric and magnetic fields |
| Answer» Correct Answer - A | |
| 503. |
Calculate the total number of electron is `1` mol of qammonia |
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Answer» Correct Answer - A::B::D One mole of `NH_(3)` contains one mole nitrogen atom and three mole hydrogen atoms .Each nitrogen atom contains electron. Therefore, number of electrons in `1` mole `NH_(3) = 7 xx 6.02 xx 10^(23) + 3 xx 6.02 xx 10^(23) = 6.02 xx 10^(24)` |
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| 504. |
The quantum number that does not describe the distance and the angular disposition of the eelctron :A. `n`B. `l`C. ` m`D. `s` |
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Answer» Correct Answer - D s-describes only spin of electron. |
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| 505. |
An electron jumps from an outer orbit to an inner orbit with the energy difference of 3.0 eV. What will be the wvelength of the line and in what region does the emission take place? |
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Answer» Correct Answer - `lamda-4140"Ã…; viible region"` `1eV=1.6xx10^(-12)erg` |
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| 506. |
The magnitude of an orbital angular momentum vector of an electron is `sqrt(6) (h)/(2pi)` into how many components will the vector split if an external field is applied to it ?A. 3B. 5C. 7D. 10 |
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Answer» Correct Answer - B 5 |
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| 507. |
The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV` The potential energy value (s) of excited state(s) for the electron in the Bohr orbit of hydrogen is(are)A. `- 3.4 eV`B. `4.2 eV`C. `-6.8 eV`D. `+ 6.8 eV` |
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Answer» Correct Answer - A Eecited state is given as `= (-13.6 eV)/(n^(2))` `e.g. n = 2 ,E = (-13.6)/(4) = - 13.6 eV` |
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| 508. |
For `d` electron, the orbital angular momentum isA. `sqrt(6) ((h)/(2pi))`B. `sqrt(2)((h)/(2pi))`C. `((h)/(2pi))`D. `2((h)/(2pi))` |
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Answer» Correct Answer - A |
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| 509. |
The position of both an electron and a helium atom is known within `1.0 nm` and the momentum of the electron is known within `50 xx 10^-26 kg ms^-1`. The minimum uncertainty in the measurement of the momentum of the helium atom is.A. `50 kg ms^-1`B. `60 kg ms^-1`C. `80 xx 10^-26 kg ms^-1`D. `50 xx 10^-26 kg ms^-1` |
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Answer» Correct Answer - D (d) The product of uncertainties in the position and the momentum of a sub atomic particle `= h//4 pi`. Since `Delta x` is same for electron and helium so `Delta p` must be same foe both the particle, i.e., `50 xx 10^-26 kg ms^-1` (given). |
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| 510. |
The position of both an electron and a helium atom is known within `1.0 nm` and the momentum of the electron is known within `50 xx 10^-26 kg ms^-1`. The minimum uncertainty in the measurement of the momentum of the helium atom is.A. `50 "kg ms^(-1)`B. `60 "kg ms^(-1)`C. `80xx10^(-26) "kg ms"^(-1)`D. `50xx10^(-26) "kg ms"^(-1)` |
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Answer» Correct Answer - D The product of uncertainties in the position and the momentum of a sub atomic particle =`h//4pi`. Since `Deltax` is same for electron and helium so `Deltap` must be same for both the particle i.e. `50xx10^(-26) kg ms^(-1)` (given) |
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| 511. |
The ratio of the energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation isA. `1//4`B. `4`C. `1//2`D. `2` |
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Answer» Correct Answer - D `E=(hc)/lambda, lambda_1=2000 Å, lambda_2=4000 Å`, so `E_1/E_2 =lambda_2/lambda_1=4000/2000=2` |
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| 512. |
The ratio of the energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation isA. `1//4`B. 4C. `1//2`D. 2 |
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Answer» Correct Answer - D `E_(2)/E_(1)=lambda_(1)/(lambda_(2))=(4000)/2000=2` |
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| 513. |
Which is not in accordance to Aufbau principleA. B. C. D. |
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Answer» Correct Answer - C Aufbau principle states that in the ground state of an atom . The orbital with lower energy is filled up first before the filling of the orbitals with a higher energy commences. Increasing order of energy of various orbitals is 1s, 2s, 2p, 3s, 4s, 3d, 3p, 5s .... etc |
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| 514. |
One of the spectral lines of cesium has a wavelength of `456` Calculate the frequency of this line |
| Answer» `v = (c )/(v) = (3 xx 10^(8) m s^(-1))/(456 xx 10^(-9)) = 6.58 xx 10^(14)s^(-1)` | |
| 515. |
In a photoelectric effect experiment irradiation of a metal with light of frequency `5.2 xx 10^(14) s^(-1)` yields elctrons with maximum kinetic ennergy `1.3 xx 10^(-19) J`.Calculate the threshold frequency `(v_(0))` for the metal |
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Answer» We know that `hv = hv_(0)+ KE` or `v_(0) = v - (KE)/(h)` `KE = 1.3 xx 10^(-19)J, v = 5.2 xx 10^(14)s^(-1), h = 6.626 xx 10^(-34) Js` :. Theshold frequency `v_(0) = 5.2 xx 10^(24)s^(-1) - (1.3 xx 10^(19)J)/(6.626 xx 10^(-34)Js)` `= 5.2 xx 10^(14)s^(-1) - 1.96 xx 10^(14)s^(-1)` `3.24 xx 10^(14) s^(-1)` |
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| 516. |
Which of the following options does not represent ground state electronic configuration of an atom?A. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(8)4s^(2)`B. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(9)4s^(2)`C. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(1)`D. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` |
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Answer» Correct Answer - B Correct configuration should be `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(1)` for the copper which has atomic number `29(._(29)Cu)`. Due to extra stability of full filled orbital of d-subshell, the last electron enter into d-orbital instead of sorbital. |
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| 517. |
Wave properties are only important for particoles havingA. High mass and low velocitiesB. Low mass and no velocityC. High mass and high velocitiesD. Low mass and high velocities |
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Answer» Correct Answer - D de-brogle theory applicable fof very small size particle |
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| 518. |
" Positronium " is the name given to an aotm like combination formed between :A. A positiron and a protonB. A positron and a neutronC. A positron and alpha-particleD. A positron an electron |
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Answer» Correct Answer - D Positron + Electron `to` Positronium. |
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| 519. |
Is the angular momentum of an electron in an atom quantized ? Explain |
| Answer» Correct Answer - Yes | |
| 520. |
A radin station is broadcasting programmes as `100` MHz frequency if the distance between the ratio station and the receved set is `300` km how long the signal would take in reach the set from the radiostating ? Also calculate the wavelength and wavelength of radio waves |
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Answer» All electromagnetic waves travel in wacuum or air with the same speed of `3 xx 10^(6) m s^(-1)` Time taken by wave to reach the reciver `= (Destance)/(Time) ` `= (300 xx 1000m)/(3 xx 10^(8) ms^(-1)) = 1 xx 10^(-8)s ` Wavelength of radition `lambda = (c )/(v) = (3 xx 10^(8) m s^(-1))/(100 xx 10^(6) s^(-1)) = 3 m` Wavelength `bar v = (1)/(lambda)= (1)/(3) = 0.33 m^(-1)` |
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| 521. |
For d subshell, the azimuthal quantum number is |
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Answer» Correct Answer - C For d orbital l=2 |
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| 522. |
What is the significance of ` psi_(210` ? Find out angular momentum , spherical nodes and angular node for ` psi_(210`. |
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Answer» ` psi` reperesnts and orbittal an ` psi_( 210) ` has ` n= 2, l = 1, m =0 i.e., 2 p` subshell . The angualr momentum ` = h/( 2 pi) sqrt ( l( l + 1))` ` (h) /(2pi) xx sqrt 2= h/( sqrt 2pi)` Also spherical node in ` 2 p = n -l - 1 = 2 - 1-2 =0` Angular node ` = l=1` . |
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| 523. |
In `psi_(321)` the sum of angular momentum, spherical nodes and angular node is:A. `(sqrt(6)h+4pi)/(2pi)`B. `(sqrt(6h))/(2pi)+3`C. `(sqrt(6h)+2pi)/(2pi)`D. `(sqrt(6h)+8pi)/(2pi)` |
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Answer» Correct Answer - A `psi_(321): n = 3, i=2, m= 1` Angular momentum `=(h)/(2pi) sqrt(l(l+1)) = (sqrt(6)h)/(2pi)` Spherical nodes `= 3 - 2 -1 = 0`, Angular node `=2` Sum of all the above `=(sqrt(6)h)/(2pi) +2 = (sqrt(6h)+4pi)/(2pi)` |
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| 524. |
Which among the folloiwing series is obtained in both absorption and emission spectrums?A. Lyman seriesB. Balmer seriesC. Paschen seriesD. Brackett series |
| Answer» Correct Answer - A | |
| 525. |
The maximum kinetic energy of photoelectrons is directly proportional to . . . Of the incident radiation. The missing word can be:A. intenstiyB. wavelengthC. wave numberD. frequency. |
| Answer» Correct Answer - C::D | |
| 526. |
Which of the following orbital(s) lie in the xy-plane?A. `d_(x^(2)-y^(2))`B. `d_(xy)`C. `d_(xz)`D. `d_(yz)` |
| Answer» Correct Answer - A::B | |
| 527. |
In which of the following sets of orbitals, electrons have equal orbital angular momentum?A. 1s and 2sB. 2s and 2pC. 2p and 3pD. 3p and 3d |
| Answer» Correct Answer - A::C | |
| 528. |
For a shell of principal quantum number n=4, there are:A. 16orbitalsB. 4 subshellsC. 32 electrons (maximum)D. 4 electrons with l=3 |
| Answer» Correct Answer - A::B::C | |
| 529. |
How many d-electrons in `Cu^(+)(At.No =29)` can have the spin quantum `(-(1)/(2))`? |
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Answer» Correct Answer - 5 Configuration is `[Ar] 3d^(10)` |
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| 530. |
The azimuthal quantum number of a non-directional orbital is |
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Answer» Correct Answer - A s orbital |
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| 531. |
According to qauntum mechanical model of H-like species, and electron can be represented by a wave function `(psi)` which contain all dynamic information about the electron. The nature of wave function depends on the type of the orbital to which the electron belongs. For an orbital `psi=[sqrt(2)/(81sqrt(3pi))]((1)/(a_(0)))^(3//2)(27-18sigma+2sigma^(2))e^((sigma)/(3))` Where, `sigma =((Zr)/(a_(0))),r` = radial distance from nucleous, `a_(0)=52.9"pm"` The number of radial and angular nodes possible for the orbital given above are respectivelyA. zero,zeroB. `0,2`C. `2,0`D. `2,1` |
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Answer» Correct Answer - C For `r=0`, if `psi(r )ne0` implies given function represents an s-orbital. From the given function total number of radial nodes obtained are `2`. |
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| 532. |
Which orbitals is non-directional ?A. `s`B. `p`C. `d`D. All |
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Answer» Correct Answer - A `B.E. = I.E.` `(I.E.)_("any atom") = (I.E.)_(H) xx ƶ^(2)` `(122.4)/(13.6) = ƶ^(2)` `ƶ^(2)=9 " "ƶ=3` `E_(2)-E_(1) = 122.4-30.6 = 91.8 eV`. |
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| 533. |
How many electron in a given atom can have the following quantum numberA. n=4,l=1B. `n=2,l=1,m=-1,s=1//2`C. `n=3`D. `n=4, l=2,m=0` |
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Answer» Correct Answer - ` (a) 6 " " (b) 1 " " (c) 18 " " (d) 2` |
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| 534. |
Rutherfords experiments , which established the nuclear model of atom , used a beam of:-A. `beta`-particles , which impinged on a mental foil and get absorbed.B. `gamma`- rays, which impinged on metal foil and ejected electron .C. Helium atoms, which impinged on a metal foil and got scattered.D. Helium nuclei, which impinged on a metal foil and got scattered. |
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Answer» Correct Answer - D `alpha`-particle (He- nucleus ) were used in Rutherford experiment . |
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| 535. |
The uncertainty in the position of an electron moving with a velocity of `1 xx 10^4 cm s^-1` (accurate up to `0.011 %`) will be :A. `1.92 cm`B. `7.68 cm`C. `0.528 cm`D. 3.8 cm |
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Answer» Correct Answer - C ( c) `Delta v =1 xx 10^4 xx (0.011)/(100) = 1.1 cm sec^-1` `Delta x = (h)/(4 pi m Delta v) = (6.626 xx 10^-27)/(4 xx 3.14 xx 9.1 xx 10^-28 xx 1.1) = 0.528 cm`. |
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| 536. |
Which of the following conditions is incorrect for a well behaved wave function `(Phi)`?A. `psi` must be single valued at any particular pointB. `psi` must be positiveC. `psi` must be a continuous function of its coordinatesD. None of the above |
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Answer» Correct Answer - B All are correct |
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| 537. |
For a d electron the orbital angular momentum isA. `2 sqrt(3) ħ`B. `ħ`C. `sqrt(6) ħ`D. `sqrt(2) ħ` |
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Answer» Correct Answer - C ( c) Angular momentum `= sqrt(l(l + 1)) ħ` For d-orbital, l = 2 Angular momentum `= sqrt(2(2 + 1)) ħ = sqrt(6) ħ`. |
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| 538. |
Calculate the wavelength of a track star running `150` metre dash in `12.1 sec` if its weight is `50 kg`.A. `9.11 xx 10^-34 m`B. `8.92 xx 10^-37 m`C. `1.12 xx 10^-45 m`D. none of these |
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Answer» Correct Answer - B (b) `lamda = h //mv = (6.626 xx 10^-24)/(50 xx (150 //12)) = 8.92 xx 10^-37 m`. |
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| 539. |
What will be the kinetic energy and total eknergy charge of an in H atom if the atom emit a photon of wavelength `4860 Å`? |
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Answer» The energy released is `E = (hc)/(lambda) = ((6.625 xx 10^(-34) Js) xx ( 3 xx 10^(8) ms^(-1)))/((4860 xx 10^(-10)m))` `= 4.09 xx 10^(-19) J` 1:. Total energy charge `= 4.09 xx 10^(-19) J` :. Total energy due to released of photon = gain in kinetic energy `= 4.09 xx 10^(-19) J` |
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| 540. |
The orbital `n = 6,l= 2 and m= 0` will be designated asA. `6d_(x^(2)`6dB. `6d_(x^(2 - y^(2)`C. `6d_(sp)`D. `6p_(2)` |
| Answer» For `6th` of energy level `1= 2` is for s sub-level and `m=1)d_(2)` orbital | |
| 541. |
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 were the hydrogen atoms travelling before collision? |
| Answer» `4.43 xx 10^4 m sec^(-1)`, | |
| 542. |
Nuclei tend to have more neutrons than protons at high mass numbers becauseA. Neutrons have neutral particleB. Neutrons have more mass than protonsC. more neutrons minimize the coulomb repulsionD. Neutrons decrease the binding energy |
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Answer» Correct Answer - C Since neutrons are uncharged but their mesonic exchanges with protons and neutrons itself contribute to stability , [It has limited application ] |
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| 543. |
Identify the CORRECT statementA. Quantum numbers (n,l,m,s) are obtained arbitrarilyB. All the Quantum numbers (n,l,m,s) for any pair of electrons in an atom can be identical under special circumstanceC. All the quantum numbers (n,l,m,s) may not be required to described an electron of an atom completelyD. All the quantum numbers (n,l,m,s) are required to described an electron of an atom completely |
| Answer» Correct Answer - D | |
| 544. |
A mixture contains `F` and `Cl` atoms . The removal of an electron form each atom of the sample requires `28 kJ` while addition of an electron to each atom of mixture releases `68.8 kJ` energy .Calcualte the % composition of mixture .Given `IE` per atoms for `F` and `Cl` are `27.91 xx10^(-22)kJ` and `20.77 xx10^(-22) kJ`. Electron gain enthaply for F and Cl are ` -5. 31 xx10 ^(-22) kJ` and `-5.78 xx 10^(22) kJ` respectivley |
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Answer» Let the mixture contain x and y mole of F and CI a. `," (x xx 6.023 xx 10^(23) mol^(-1)) (27.91 xx 10^(23) KJ) + (y + 6.023 xx 10^(23)) (20.77 xx 10^(22) KJ) = 284 KJ`…..(1) b. `(x xx 6.023 xx 10^(23))(5.23 xx 10^(-23))(y xx 6.23 xx 10^(-23)) (5.78 xx 10^(-22)KJ) = 5.78 KJ`....(ii) `:. x(1681) + y(1251) = 284` x(333.1) + y(348.1) = 68.8` Solving for x and y we get `x = 0.076,y = 0.125 ` % of `F = (0.076 xx 100)/((0.076 + 0.125)) = 37.81` `% of CI = 100 - 37.81 = 62.19` |
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| 545. |
Which of the following statements is correct ?A. Indigo light have less energy as compared to yellowB. Bright lines are formed on dark background of the photographic film in emission spectrumC. X-rays have more frequency as compared to `gamma`-raysD. Bright lines are formed on dark background of the photographic film in absorption spectrum |
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Answer» Correct Answer - B |
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| 546. |
Correct configuration of `Fe^"+3"` [26] isA. `1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^5`B. `1s^2 , 2s^2 sp^6 , 3s^2 3p^6 3d^3, 4s^2`C. `1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^6,4s^2`D. `1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^5,4s^1` |
| Answer» Correct Answer - A | |
| 547. |
The electrons identified by quantum numbers n and lA. n=4 , l=1B. n=4 , l=0C. n=3 , l=2D. n=3 , l=1 |
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Answer» Correct Answer - B (1)4p , (2)4s , (3)3d , (4)3p According to (n+l) rule , increasing order of energy (4) lt (2) lt (3) lt(1) |
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| 548. |
Calculate ratio of wavelength for a proton and `alpha-`particle . If their KE are same. |
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Answer» Correct Answer - 2 |
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| 549. |
Energy of the electron in Hydrogen atom is given byA. `E_n=-(131.28)/n^2 "kJ mol"^(-1)`B. `E_n=-(131.33)/n^2 "kJ mol"^(-1)`C. `E_n=-(1313.3)/n^2 "kJ mol"^(-1)`D. `E_n=-(131.13)/n^2 "kJ mol"^(-1)` |
| Answer» Correct Answer - C | |
| 550. |
Electrons with a kinetici energy of `6.023xx10^(4)J//mol` are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nm. The minimum amount of energy required to remove an electron fro the metal atom is:A. `2.3125 xx 10^(-19)` JB. `3 xx 10^(-19)` JC. `6.02 xx 10^(-19)` JD. `6.62 xx 10^(-34)` J |
| Answer» Correct Answer - A | |