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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.
| 2351. |
Calculate the number of elctron which will together weigh `1 g ` |
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Answer» Correct Answer - A::B::C Number of electron `= (1)/(9.1 xx 10^(-20) g ("electron")^(-1)) = 1.099 xx 10^(27) "electron" g^(-1)` |
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| 2352. |
Which experiment observation led to the following conclassions ? a. Atom contains a massive positive center Size of the nucleus is very small |
| Answer» Correct Answer - A | |
| 2353. |
Energy levels A. B, C. of a certain atom correspoinds to increasing value fo enrgy i.e., E_A lt E_B lt E_C If ` lambda _1, lambda_2`, and ` lambda_3` are the wavelengths of radistions corresponding to the transitions ( C) to ` B, B ` to (A) and ( C) to (A) respectively wich fo the following wstatemnt is correct .A. ` lambda_3 = lambda_1 + lambda_2`B. `lambda_3 ( lambda_2 )/(lambda_1+lambda_2)`C. `lambda+1+ lambda-02 + lambda_3 =0`D. `lambda_3^2 = lambda_1^2+lambda_2^2` |
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Answer» Correct Answer - B ` E_(3) = E_(1) E_(2) or ( hc)/(lambda_(3)) = (hc)/(lambda_(1)) + (hc)/(lambda_(2))`. |
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| 2354. |
If the electron enrgy is ` -3 .4` eV, find the principal quantum number fo H-atom . |
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Answer» Correct Answer - 2 `E_n = E_1/n2 ( :. E_1 "for H-atom" =- 13. 6 eV)` or ` n^2 = E_1 E_n = ( 13,6)/( -3.4) :. N=2` . |
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| 2355. |
If the total enrgy fo an electron in a hydrogen linke atom in an ecited state is ` - 3. 4 eV` , the the de-Broglie wavelngth of the electron is :A. `6.6xx10^(-10)`B. ` 3 xx 10^(10)`C. ` 5 xx 10^(-9) `D. ` 9. 3 xx 10 ^(10)` |
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Answer» Correct Answer - A Total energy ` = (-e^2)/(2r_n) =- 3 . 4 eV = E_(1) /n^2` `:. N^(2) = ( -13 . 6)/( -3 . 4) = 4 :. N=2` The velovity in II orbit ` = u_(1)/2 = ( 2 . 18 xx 10^8 )/2 cm sec ^(-1)`. ` therefore lambda = h/( mu) = (6.6 xx 10^(27)xx 2)/( 9. 108 xx 10^(-23) xx 2 . 18 xx 10^8)`. ` = 6.6. xx 10^(-10) ` ` E_(1) =- 13 . 6 eV`. |
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| 2356. |
For an atom with atomic numner `14` the number of orbits and orbitals in which elrctrons are prresent are respectively :A. ` 3,6`B. ` 6,3`C. ` 7,3`D. 3,8 |
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Answer» Correct Answer - D Electronic configuration of atomic no. `14 ` is ` 1s^(2)2 . 2s^2 2p^6 . 3s^2 3p^2` ( 3 orbits and (8) orbitals ). |
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| 2357. |
As an electron is brought form an infinite distance close to the nucleus fo the atom the enrgy of the electron -nucleus system :A. Increased to a grater positive valueB. Decreases to a smaller positiive valueC. Decreases to a smaller positiive valueD. Increased to a grater positive value |
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Answer» Correct Answer - C As a result fo attracton some enegy is released. |
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| 2358. |
Light of wavelength ` 300 xx 10^(-9)` m strikes a metal surface with photoeoctric work function of ` 2. 13 eV`. Find out the kinetic enrgy fo the most energetic photoelectron . |
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Answer» Energy of photon = Work function + Kinetic enrgy Kinetic energy = Energy of photo -Work function ` = (hc)/(lambda) - "Work function"` `(therefore "Work function" = 2 . 13 eV = 2 . 13 xx 1. 6 xx 10^(-19) J)` `=[(6.626 xx 10^(-34) xx 3.0 xx 10^8)/(300 xx 10^(-9))] - 2 . 13 xx 1. 6 xx 10^(-19)` ` 6.626 xx 10^(-19) - 3. 41 xx 10 ^(-19) ` ` K.E . = 3 . 216 xx 10^(-19) J` The is the kinetic energy of most energyetic photoelectron. |
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| 2359. |
Nucleus of an element contains 9 protons Its valency would be :A. 1B. 3C. 2D. 5 |
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Answer» Correct Answer - A Since its nucleus contain 9 proton so its. Atomic number is 9 and its electronic configuration is 2,7. So it require one more electron to complete its octet. Hence its valency is 1. |
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| 2360. |
A single electron system has ionization enrgy ` 1. 118 xx 10^7 J mol^(-1)`. Calculate the numner of protons in the nucleus fo te system . |
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Answer» Correct Answer - 3 `:.` Ionization enrgy ` = Z^2 /n^2 xx 21. 69. xx 10^(-19) J` ` :. ( 1.118 xx 10^7)/( 6.023 xx 10^(23)) = (Z^2)/((1)^2) xx 2 1. 69 xx 10^(-19)` ` :. Z=3`. |
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| 2361. |
A single electron system has ionisation energy `11180 KJ "mole"^(-1)`. The number of protons in the nucleus of the system is....... |
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Answer» Correct Answer - 6 `I.E. = (1312)/(n^(2)) xx Z^(2)` |
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| 2362. |
The first five ionization energies of an element are `801,2428,3660,25030,32835 "in" "kJ"//"mol"`. Then the element could be.A. a halogenB. a noble gasC. a third group elementD. a second group element |
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Answer» Correct Answer - C ( c) Energy difference between `3rd` and `4 th IP`s is very high. |
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| 2363. |
A single electron system has ionization energy 11180 kJ `mol^(-1)`. Find the number of protons in the nucleus of the system. |
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Answer» Correct Answer - Z=3 `IE=(Z^(2))/(n^(2))xx21.69xx10^(-19)J` `(11180xx10^(3))/(6.023xx10^(23))=(Z^(2))/(1^(2))xx21.69xx10^(-19)` `Z~~3`. |
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| 2364. |
Binding energy per nucleon of three nuclei A, B and C are 5.5, 8.5 and 7.5 respectively. Which one of the following nuclei is most stable?A. AB. BC. CD. Cannot be predicted |
| Answer» Correct Answer - C | |
| 2365. |
Which of the following set of quantum numbers is not possible for an electron in the ground state of an atom with atomic number `19`?A. n=2,l=0,m=0B. n=2,l=1,m=0C. n=3,l=1,m=-1D. n=3,l=2,m=`+-2` |
| Answer» Correct Answer - D | |
| 2366. |
Helium nucleus is composed of two protons and two neutrons if the atomic mass is `4.00388`, how much energy is released when the nucleus is constituted? (Mass of proton =1.00757, mass of neturon=1.00893)A. 283MeVB. 28.3 mEvC. 2830 MeVD. 2.83 MeV |
| Answer» Correct Answer - B | |
| 2367. |
Total number of electrons, protons and neutrons present in the nucleus of `._(92)U^(238) `isA. e=92,p=92,n=146B. e=92,p=92,n=148C. e=0,p=92,n=146D. None of these |
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Answer» Correct Answer - e=0,p=92,n=146 There is no electron in the nucleus. `n_(e)=0,n_(p)=92,n_(n)=238-92=146` |
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| 2368. |
An isotone of `._(32)^(76)Ge` is-(a) `._(32)^(77)Ge`(b)`._(33)^(77)As`(c)`._(34)^(77)Se`(d)`._(34)^(78)Se`A. `._(32)^(77)Ge`B. `._(33)^(78)As`C. `._(34)^(77)Se`D. `._(34)^(78)Se` |
| Answer» Correct Answer - D | |
| 2369. |
Electron in a sample of `H-` atoms make transitions from state `n=x` to some lower excited state. The emission spectrum from the sample is found to contain only the lines belonging to a particular series. If one of the photons had an energy of `0.6375 eV`. Then find the value of `x. [" Take " 0.6375eV=(3)/(4)xx0.85eV]`.A. 16B. 24C. 8D. 20 |
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Answer» Correct Answer - 3 We have, `" "DeltaE=(3)/(4)xx0.85eV` as energy `=0.6375` the photon will belong to brakett series ( as for brackett `0.31 le E le0.85)` `0.85xx(1-(1)/(4))=13.6((1)/(4^(2))-(1)/(n^(2)))` `0.85(1-(1)/(4))=(13.5)/(16)[1-((4)/(n))^(2)]" ":. (4)/(n)=(1)/(2)" "rArr" "n=8` |
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| 2370. |
What a certain was metal was irradiatiobn with light of frequency `1.6 xx 10^(16) Hz` the photoelectron emitted but the kinetic energy as the photoelectron emitted when the same metal was irradiation with light of frequency `1.0 xx 10^(16) Hz` .Calculate the threslold frequency `(v_(0))` for the metal |
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Answer» `hv = hv_(0) = KE` `KE_(1) - h(v - v_(0))` `KE_(2) = h(v_(2) - v_(0)) = (KE_(1))/(2)` `:. (v_(2) - v_(v_(0)))/(v_(1) - v_(0)) = (1)/(2) rArr (1.0 xx 10^(16) - v_(0))/(1.6 xx 10^(16) - v_(0))m = (1)/(2)` `rArr v_(0) = 4 xx 10^(15)Hz` |
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| 2371. |
Gives the values of all the four quantum numbers for `2p` electron in nitrogen `(Z= 7)` |
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Answer» The electronic configuraqtion of nitrogen atom is `1s^(2)2s^(2)2p_(2)^(1) 2p_(1)^(2) 2p_(4)^(3)` For the `2p` elementss `n= 2,l = 1, m = -1,o= 1, x = + 1//2 or -1//2` (All `2p` electron have parallel spin ) |
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| 2372. |
The angular momentum of p electron isA. `sqrt(6). (h)/(2 pi)`B. `sqrt(3) (h)/(2 pi)`C. `sqrt((3)/(2)) (h)/(pi)`D. `(h)/(sqrt(4) pi)` |
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Answer» Correct Answer - D (d) Orbital angular momentum `= sqrt(l(l + 1)) xx (h)/(2 pi)` For `p-"electron" (l = 1)` =`sqrt(1(l + 1)) xx (h)/(2 pi) = sqrt(2) xx (h)/(2 pi) = (h)/(sqrt(2) pi)`. |
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| 2373. |
What is the maximum number of elements if the electrons above `n = 4` do not exist in nature ?A. 40B. 40C. 44D. 100 |
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Answer» Since `n = 1,2,3, and 4` `(1s)/(2),(2s 2p)/(8),(3s 3p 3d)/(18) ,(4s 4p 4d 4f)/(32)` Thus total number of existent element is `2+8+18+32 = 60` |
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| 2374. |
Minimum de-Broglie wavelength is associated with.A. ElectronB. ProtonC. `CO_2` moleculeD. `SO_2` molecule |
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Answer» Correct Answer - D (d) `lamda = (h)/(mv)`. For same velocity `lamda prop (1)/(m)`. `SO_2` molecule has least wavelength because their molecular mass is high. |
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| 2375. |
The sub-shell `2d` is not posible becauseA. `n = 1`B. `l gt n`C. `n lt l`D. None of these |
| Answer» For sub-shell `2d, n = 2 and l = 2 ` and the value of n and l can never be equal | |
| 2376. |
What is the maximum number of electron in the posible sub-shell for `n + 1 = 4` ?A. 8B. 6C. 12D. 16 |
| Answer» `{:(n+l=4),(4+0=4s to 2),(3+1=3p to 6),(bar("Total 8")):}` Maximum number of electrons | |
| 2377. |
if Helium atom and Hydrogen molecule are moving with the same velocity , their wavelength ratio will beA. `4:1`B. `1:2:`C. `2:1`D. `1:4` |
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Answer» Correct Answer - B `lambdaprop1/m=lambda_"He"/lambda_(H_2) =m_(H_2)/m_(He) rArr lambda_(He)/lambda_(H_2)=2/4 =1:2` |
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| 2378. |
Which of the following statement are correct ?A. The electronic configuration of Cr is `[Ar]3d^(5)4s^(1)`(atomic number of Cr is `24)`B. The magnetic quantum number may have a negative valueC. In silver atom `23` electron have spin of one type and `24` of the opposite type (atomic number of `Ag is 47)`D. The oxidation state of nitrogen in `HN_(3)` is `-3` |
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Answer» Correct Answer - A::B::C `Cr(Z = 24)` `[Ar]3d^(6)4s^(1)` This is because `d^(5)` is a more half -filled configuration Reason for the stablity of half-filled and fallty filled ortained are symmetry and excharge energy For every value of l m can be fromn -l to + l through 0, `Ag(z = 47)` The oxidation state of N in `HN_(3)` is not `- 3`but `- 1//3` |
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| 2379. |
The principal quantum number of an atom is related in theA. Size of the orbitB. Spin angular momentumC. orbital angular momentumD. Orientation of the orbital in space |
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Answer» Correct Answer - A The principal quantum number (n) is related to the size to the orbital (n=1,2,3,…) |
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| 2380. |
The principal quantum number of an atom is related in theA. Size of the orbitalB. Spin angular momentumC. Orbital angular momentumD. Orientation of the orbit in space |
| Answer» `{:(n + 1 = 7),(7 + 0 = 7s),(6 + 1 = 6p),(5 + 2 = 5d),(4 + 3 = 4f):}` | |
| 2381. |
A dust particle has mass equal to `10^(-11)g` , diameter equal to `10^(-4)` cm and velocity equal to`10^(-4) cms^(-1)`. The error in the measurment of velocity is `0.1%` . Calculate the uncertiainty in its position. Comment on the result |
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Answer» ` u = 10 ^(-4) "cm sec"^(-1)` ` :. Delta u = ( 0. 1 xx 10^(-4))/(100) = 1xx 10^(-7) cm sec^(-1)` Now ,` Delta u . Delta x = h/( 4 pi m)` ` :. Delta x = ( 6.626 xx 10^(-27))/( 4 xx 3. 14 xx 10^(-11) xx 10^(-7))` `=0.527 xx 10^(-9)` cm The uncertainty in position as compared to particle ` = (Deltax )/( diameter )` ` = ( 0. 527 xx 10^(-9))/( 10^4) = 0. 527 xx 10^(-5)` The factor is very small and alost negligible for microscopic particels , |
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| 2382. |
Which electronic level would allow the hydrogen atom to absorbe a photon but not to emit a photon ?A. 3sB. 2pC. 2sD. 1s |
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Answer» Correct Answer - D The lowest energy state is 1s it is not posible from this state to lost energy |
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| 2383. |
The principal quantum number of an atom is related in theA. Size of the orbitalB. Spin angular momentumC. Orienitation of the orbital in spaceD. Orbital abgular momebntum |
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Answer» Correct Answer - A The large the value of the principle quantum number, large the size of th eshell and hence the orbital |
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| 2384. |
Point out the followings: (a) How many energy subshells are pssible in ` n=3` level ? (b) How many orbitals of all kinds are possible in ` n=3` level ? |
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Answer» (a) No. of subshells in a shell = no . Of shells = 3 (b) No. of orbitals = `( "no. of shell")^2 = 3^2 =9`. |
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| 2385. |
Which of the following sets of quantum number is restrictedA. n=3, l=1, m=+2B. n=3, l=2, m=-2C. n=3, l=1, m=+1D. n=3, l=1, m=-1 |
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Answer» Correct Answer - A For n=3, l=1, m=-1,0,+1 |
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| 2386. |
For the n=2 energy level, how many orbitals of all kinds are possibleA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - C For the n=2, energy level orbitals of all kinds are possible `n^2 , 2^2 =4` |
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| 2387. |
The velocity of an electron in the first Bohr orbit of hydrogen atom is `2.19 xx 10^(6)ms^(-1)`. Its velocity in the second orbit would beA. `1.10 xx 10^(6)ms^(-1)`B. `4.38 xx 10^(6)ms^(-1)`C. `5.5 xx 10^(5)ms^(-1)`D. `8.76 xx 10^(6)ms^(-1)` |
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Answer» Correct Answer - A `v_(n) = (v_(1))/(n)` |
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| 2388. |
Energy of electron moving in the second orbit of `He^(+)` ion isA. `-13.6ev`B. `-3.4ev`C. `-1.51 ev`D. `-0.84ev` |
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Answer» Correct Answer - A `E = (-13.6)/(n^(2)) xx Z^(2)` |
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| 2389. |
A dispositive ion `16` protons what is the number of el,ectron is its tertpositive ion?A. `16`B. `14`C. `12`D. `10` |
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Answer» `X^(2+)` has `16` protons then in `X`, there are `16` protons and `16` electrons in `X^(2+)`, there are `16` protons and `14` electrons in `X^(4+)`, there are `16` protons and `12` electrons |
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| 2390. |
Consider the following nuclear reactions involving X and Y. `X rarr Y + ._(2)^(4)He` `Y rarr ._(8)O^(18) + ._(1)H^(1)` If both neutrons as well as protons in both the sides are conversed in nuclear reaction then moles of neutrons in 4.6 g on X :A. `2.4 N_(A)`B. `2.4`C. `4.6`D. `0.2 N_(A)` |
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Answer» Correct Answer - B |
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| 2391. |
Write the complete symbol a. The nucleus with atomic number `16` and mass number `82` b. The nucleus with atomic number `35` and mass number `80` c. The nucleus with atomic number `4` and mass number `9` |
| Answer» a.`._(16)X^(32)` or `._(16)S^(32)` b. `._(35)X^(80)` or `._(35)Br^(80)`c.`_(4)X^(9)` or `._(4)Be^(9)` | |
| 2392. |
Consider the following nuclear reactions involving X and Y. `X rarr Y + ._(2)^(4)He` `Y rarr ._(8)O^(18) + ._(1)H^(1)` If both neutrons as well as protons in both the sides are conversed in nuclear reaction then moles of neutrons in 4.6 g on X :A. `3,2.4N_(A)`B. `3,2.4`C. `2,4.6`D. `3,0.2N_(A)` |
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Answer» Correct Answer - B `E = (hc)/(lambda) implies (1240)/(lambda_(nm))ev " "E=(1240)/(31)implies 40 eV` `40=12.8 +K.E.` `K.E. = 10-12.8=27.2eV` `K.E. = 27.2 xx1.6xx10^(-19)` `27.2 xx1.6xx10^(-19)=(1)/(2)xx9.1xx10^(-31)xxv^(2)` `v = 2.18 xx sqrt(2) xx 10^(6) m//s` |
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| 2393. |
Energy of atomic orbitals in a particular shell is in the orderA. s lt p lt dlt fB. s gt p gt d gt fC. p lt d lt flt sD. f gt dgt s gt p |
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Answer» Correct Answer - A In particular shell, the energy of atomic orbital increases with the value of l |
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| 2394. |
The quantum numbers n=2, l=1 representA. 1s orbitalB. 2s orbitalC. 2p orbitalD. 3d orbital |
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Answer» Correct Answer - C l=1 is for p orbital |
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| 2395. |
Calculate ratio of wavelenght of limiting line of Blamer and pashcen series for an atom. |
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Answer» Correct Answer - `4//9` |
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| 2396. |
If the atomic weight of `Zn` in `70` and its atomic number in `30`, Then what be the atomic weight of `Zn^(2+)`? |
| Answer» In the formation of `Zn^(2+)` from Zn ,two electrons are removed .The number of proton and neutrons remain unchanged hence , the atomic weight of `Zn^(2+)` will be `70` | |
| 2397. |
Statement-1: K.E. of photoelectrons increases with increases in frequency of incident ligh `(VgtV_(0)).` Statement -2: Whenever intensity of light is increases the magnitude of photocurrent always increases. |
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Answer» Correct Answer - c |
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| 2398. |
The mass numbers of three istopes of an element are `10,12 and 14` units .Their precentage abundance is `80,15 and 5` respectively .What is the atomic weight of the element ? |
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Answer» The precentage abundences of isotopes are `80,15, and 5` .This is `16:3:1` Total of the ratio `= 16 + 3 + 1 = 20` `"Average weight" = (% "abundance" xx" atomic weight")/("Total of ratio" )` `"Average weight" = (10 xx 16 + 12 xx 3 + 14 xx 1)/(20) = 10.5` Thus the average atomic weight is `10.5` unit. |
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| 2399. |
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 emmision will occurB. `2.18xx10^(6)m//s`C. `2.18sqrt(2)xx10^(6)m//s`D. `8.72xx10^(6)m//s` |
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Answer» Correct Answer - C Frequency `=(1)/(T) prop (v)/(T) prop (ƶ//n)/(n^(2)//ƶ)` Frequency `=(1)/(T) prop (ƶ^(2))/(n^(3))" "Tprop(n^(3))/(ƶ^(2))=(1//4)/(8//1) =(1)/(32)` |
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| 2400. |
The species which has its fifth ionization potential equal to 340 V is :A. `B^(+)`B. `C^(+)`C. `B `D. `C ` |
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Answer» Correct Answer - C |
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