InterviewSolution
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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.
| 1001. |
A hydrogen like atom (atomic number Z) is in a higher excited satte of quantum number n .This excited atom can make a transition to the first excited state by succesively emitting two photon of energies `10.20 eV` and `17.00 eV` .Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting twio photon of energy `4.25 ev` and `5.95 eV` Determine the followings:The hydrogen -like atom in the question isA. `Li^(2+)`B. `He^(Theta)`C. HD. None |
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Answer» Correct Answer - A `Li^(2+)` |
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| 1002. |
A hydrogen like atom (atomic number Z) is in a higher excited satte of quantum number n .This excited atom can make a transition to the first excited state by succesively emitting two photon of energies `10.20 eV` and `17.00 eV` .Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting twio photon of energy `4.25 ev` and `5.95 eV` Determine the followings: The atom during transition from `n = 1` to `n = 2` emit radiation in the region ofA. VisibleB. Infira-redC. UVD. None |
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Answer» Correct Answer - A Visible |
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| 1003. |
A hydrogen like atom (atomic number Z) is in a higher excited satte of quantum number n .This excited atom can make a transition to the first excited state by succesively emitting two photon of energies `10.20 eV` and `17.00 eV` .Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting twio photon of energy `4.25 ev` and `5.95 eV` Determine the followings: The excited sate (n) of the atom isA. 4B. 6C. 8D. 3 |
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Answer» Correct Answer - B `n= 6` `n rarr 2: Delta E = 10.2 + 17 = 27.2 eV` `n rarr 3: Delta E = 4.25 + 5.95 = 10.2 eV` `{:(implies 27.2 = 13.6 Z^(2)(1/(2^(2))-1/(n^(2)))),(and 10.2 =13.6Z^(2)(1/(3^(2))-1/(n^(2)))):}} Rightarrow `Z = 3` and ` n =- 6` |
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| 1004. |
Which of the following is true ?A. The outer electronic configuration of the ground state chromium atom is `3d^(4)4s^(2)`B. Gamma rays are electroomagnetic radiations of wavelength of `10^(-6) cm"to" 10^(-5) cm `C. The energy of the electron in the `3d` orbital is less than that in the `4s` orbital of a hydrogen atomD. The electron density in the xy plane in `3d_(s^(2) - y^(2)` orbital is zero |
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Answer» Correct Answer - C a. The configuration is `3d^(5) 4s&^(2)` b The wavelength atom , the enrgy of an order of ` 10^(-11) m ` c. In hydrogen atom , the energy of an el,ectron depends only on the principle quantum number of the orbitals which it occupes d. In xz plane is not electron density if an electron occupies `3d_(x^(2) - y^(2)` orbital |
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| 1005. |
Compared with an atom of atomic weight `12` and atomic number `6`, the atom of atomic weight `13` and atomic number `6`.A. Contains more neutronsB. Contains more electronsC. Contains more protonsD. Is a different element |
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Answer» Correct Answer - A (a) `._6A^12` and `._6 X^13` both are isotopes but have different no. of neutrons. `._6 A^12, A "have" p = 6, e = 6 "and" n = 6` and `._6X^13, X "have" p = 6, e = 6 "and" n = 7`. |
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| 1006. |
What is the final velocity of an electron accelerating through a potential of `1600 V` if its initial velocity is zero . |
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Answer» Correct Answer - A::B::C `(1)/(2) mv^(2) = eV` `v = sqrt((2eV)/(m)) = [(2 xx 1.6 xx 10^(34) xx 1600)/(9.1 xx 10^(-31))] ` `= 2.37 xx 10^(7) m s^(-1)` |
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| 1007. |
Calculate the de Broglie wavelength for a beam of electron whose energy is `100 eV` |
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Answer» Correct Answer - A::B::C::D `E = (1)/(2) mv^(2) = 100 eV = 100 xx 1.6 xx 10^(-19) J` `v^(2)= (2E)/(m)` `v = ((2E)/(m))^(1//2)` `lambda = ((2E)/(m))^(1//2) = sqrt(2mE)` `lambda = (h)/(mv) = (h)/(sqrt(2mE)0) meter ` `lambda = (6.6 xx 10^(-34))/((2 xx 9.1 xx 10^(-31) xx 100 xx 1.6 xx 10^(-19))^(1//2))` `= 1.23 xx 10^(-10) m = 1.23 Å` |
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| 1008. |
The ratio of kinetic energy and potential energy of an electron in a Bohr of a hydrogen -like species isA. `1//2`B. `-1//2`C. `1`D. `-1` |
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Answer» Correct Answer - B `[{:(PE= - (Ze^(2))/(r_(n))","KE = (1)/(2)(Ze^(2))/(r_(n))),(E_(Total) = (-Ze^(2))/(r_(n)) + (1)/(2)(Ze^(2))/(r_(n)) = (-Ze^(2))/(r_(n))),(E_(Total) = (1)/(2) PE),(KE = - (1)/(2)PE):}]` `(KE)/(PE) = (1)/(2)(Ze^(2))/(r_(n))//(-Ze^(2))/(2r_(n)) = - (1)/(2)` |
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| 1009. |
The ratio of kinetic energy and total energy of an electron in a Bohr of a hydrogen -like species isA. `1//2`B. `-1//2`C. `1`D. `-1` |
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Answer» Correct Answer - D `(KE)/(E_(Total)) = (1)/(2)(-Ze^(2))/(r_(n))//(-Ze^(2))/(2r_(n)) = - 1` |
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| 1010. |
Band gap in germanium is small. The energy spread of each germanium atomic energy level is infinitesimally small.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. |
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Answer» Correct Answer - A (a) Germanium, semi-conductor substance, has small hand gap in comparison to insulator (non-metal). |
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| 1011. |
The ratio of potential energy and total energy of an electron in a Bohr of a hydrogen -like species isA. 2B. -2C. 1D. -1 |
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Answer» Correct Answer - A `(KE)/(E_(Total)) =(-Ze^(2))/(r_(n))//(-Ze^(2))/(2r_(n)) = 2` |
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| 1012. |
If `12.0 g` body is traveling along the x-axes at `100 cms^(-1)` within `1 cm s ^(-1)` .What is the uncertainty in its position ? |
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Answer» Correct Answer - A::C::D The velocity has uncertainly of `2 cm s^(-1)(99 "to cm" s^(-1))` `Delta x = (h)/(4pi m Delta x) = (6.63 xx 10^(-34) erg s)/(4 xx 3.14 xx 1 g xx 2 cm s^(-1)) = 3 xx 10^(-28) cm = 3 xx 10^(-28) m ` `Delta x = (6.63 xx 10^(-34)J s)/(4 xx 3.14 xx 9.1 xx 10^(-31) kg xx 2 m s^(-1)) = 3 xx 10^(-5) m = 30 mu m ` |
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| 1013. |
For which of the following sets of quantum numbers, an electrons will have the highest energy ?A. `{:(,"n","l","m"," "s),(,3,2,1," "(1)/(2)):}`B. `{:(,"n","l","m"," "s),(,4,2,-1," "(1)/(2)):}`C. `{:(,"n","l","m"," "s),(,4,1,0,-(1)/(2)):}`D. `{:(,"n","l","m"," "s),(,5,0,0,-(1)/(2)):}` |
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Answer» Correct Answer - B (b) For n=3, l=2 the subshell is 3d(n+l=5) n=4, l=2 the subshell is 4d(n+l=6) n=4, l=1 the subshell is 4dp(n+1)=5 n=5, l=0, the subshell is 5s(n+l=5) According to (n+l) rule greater the (n+l) value , greater the energy that is 6. |
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| 1014. |
Band gap in germanium is small. The energy spread of each germanium atomic energy level is infinitesimally small.A. Statement 1 is true , statement 2 is true , statement 2 is a correct explanation for statement 1B. Statement 1 is true , statement 2 is true , statement 2 is not a correct explanation for statement 1C. Statement 1 is true , statement 2 is falseD. Statement 1 is false , statement 2 is true |
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Answer» Correct Answer - C Germanium , semi-conductor substance , has small band gap in comparison to insulator (non-metal) . On moving top to bottom in the group 14 band gap decreases. |
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| 1015. |
The decrerasing order of energy for the electrons represented by the following sets of quantum number is : 1.`n = 4,l = 0,m = 0,s = +- 1//2`2.`n = 3,l = 1,m = 1,s = - 1//2`3.`n = 3,l = 2,m = 0,s = + 1//2`4.`n = 3,l = 0,m = 0,s = - 1//2`A. `1 gt 2 gt 3 gt 4`B. `2 gt 1 gt 3 gt 4`C. `3 gt 1 gt 2 gt 4`D. `4 gt 3 gt 2 gt 1` |
| Answer» Correct Answer - C | |
| 1016. |
The orbital with zero orbital angular momentum is.A. `s`B. `p`C. `d`D. `f` |
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Answer» Correct Answer - A Orbital angular momentum `=sqrt(l(l+1)).(h)/(2pi)` for `l=0` |
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| 1017. |
The decreasing order of energy for the electrons represented by the following sets of quantum number is : 1.`n = 4,l = 0,m = 0,s = +- 1//2`2.`n = 3,l = 1,m = 1,s = - 1//2`3.`n = 3,l = 2,m = 0,s = + 1//2`4.`n = 3,l = 0,m = 0,s = - 1//2` |
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Answer» Find `( n + l)` for each set ` For (1) ( n + l) = 4 i.e., 4s` ` For (2) ( n + l) = 4 i., e , 3 p` ` For (3) ( n+ l) =5 i.e, 3d` ` For (4) ( n + l) =3 i.e, 3s` (1) Lower is the value of `( n + l)` lower is energy level . (2) If `( n +l)` are same then orbital with lower values of ( n) possesses lower energy . `:.` Decreasing ofer of energy ` 3 gt 1 gt 2 gt 4 .` |
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| 1018. |
The angular momentum of an electron is zero. In which orbital may it be present?A. 2sB. 2pC. 3dD. 4f |
| Answer» Correct Answer - A | |
| 1019. |
The magnetic quantum number specifies.A. Size of orbitalsB. Shape of orbitalsC. Orientation of orbitalsD. Nuclear stability |
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Answer» Correct Answer - C The magnetic quantum number specifies orientation of orbitals |
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| 1020. |
The work function `(phi)` of some metals is listed can have principal quantum of metals which will show photoelectric effect when light of `300` nm wavelength falls on the metal is Metal `Li Na K Mg Cu Ag Fe Pt W phi(eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75 ` |
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Answer» Correct Answer - D Since the value of `hc = 1240 eV nm ` Energy coreresponding to `lambda = 300 nm`is `- (1240 eV nm)/(100 nm) = 4.134 eV` Electron having `phi lt 4.134 eV ` will show photoelectric effect, are Li ,Ni, K and` Mg rArr` So number of means showing photo-electric will be four ,i.e. Li ,Ni, K and Mg |
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| 1021. |
The de-Broglie wavelength of a tennis ball mass ` 60 g` moving with a velocity of ` 10 m` per second is approximately :A. `10^-16 m`B. `10^-25 m`C. `10^-33 m`D. `10^-31 m` |
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Answer» Correct Answer - C ( c) `lamda = (h)/(mv) = (6.626 xx 10^-34)/(60 xx 10^-3 xx 10) = 10^-33`. |
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| 1022. |
An electron is moving with a kinetic energy of `4.55 xx 10^-25 J`. What will be Broglie wavelength for this electron ?A. `5.28 xx 10^-7 m`B. `7.28 xx 10^-7 m`C. `2 xx 10^-10 m`D. `3 xx 10^-5 m` |
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Answer» Correct Answer - B (b) `KE = (1)/(2) mv^2 = 4.55 xx 10^-25` `V^2 = (2 xx 4.55 xx 10^-25)/(9.1 xx 10^-31) = 1 xx 10^6` `V = 10^3 m//s` de Broglie wavelength `lamda = (h)/(mv)` =`(6.626 xx 10^-34)/(9.1 xx 10^-31 xx 10^3) = 7.28 xx 10^-7 m`. |
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| 1023. |
Statement : `._(24)Cr` has more paramangetic nature than ` ._(25)Mn`. Explanation : `Cr` has more number of unpaired electron than Mn.A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - C `._(24)Cr` has all six unpaired electrons whereas `._(25)Mn` had five unpaired out of seven valency electron . Both have ` 5d ` electrons . |
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| 1024. |
The quantum number which specifies the location of an electron as well as energy isA. Principal quantum numberB. Azimuthal quantum numberC. Spin quantum numberD. magnetic quantum number |
| Answer» Correct Answer - A | |
| 1025. |
Which of the following sequence is correct as per Aufbau principle ?A. 3s lt 3d lt 4s lt 4pB. 1s lt 2p lt 4s lt 3dC. 2s lt 5s lt 4p lt 5dD. 2s lt 2p lt 3d lt 3p |
| Answer» Correct Answer - B | |
| 1026. |
The energy of an electron of `2p_y` orbital isA. Greater than that of `2p_x` orbitalB. less than that of `2p_x` orbitalC. Equal to that of 2s orbitalD. Same as that of `2p_z` orbital |
| Answer» Correct Answer - D | |
| 1027. |
The hydrogen line spectrum provides evidence for theA. Heisenberg uncertainty principleB. Wave-like properties of lightC. Diatomic nature of `H_(2)`D. Quantized nature of atomic energy states |
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Answer» Correct Answer - D Atomic orbits have fixied energies |
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| 1028. |
An `alpha` particle of momentum p is bombarded on the nucleus, the distance of the closest approach is r, if the momentum of `alpha`-particle is made to 6p, then the distance of the closest approach becomesA. `4r`B. `2r`C. `16r`D. `(r)/(36)` |
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Answer» Correct Answer - D `(p^(2))/(2m) =(q_(1)q_(2))/(r)` at closes approach |
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| 1029. |
Statement-I : The ground state configuration of Cr is `3d^(5) 4s^(1)`. Because Statement-II : A set of exactly half filled orbitals containing parallel spin arrangement provide extra stability.A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-I.B. Statement-I is true, Statement-II is true , Statement-II is NOT a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - A | |
| 1030. |
Statement-I : The ground state configuration of Cr is `3d^(5) 4s^(1)`. Because Statement-II : A set of exactly half filled orbitals containing parallel spin arrangement provide extra stability.A. If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-4B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-3C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - A Electronic configuration concept |
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| 1031. |
The orbital configuration of `._(24)Cr` is `3d^(5) 4s^(1)`. The number of unpaired electrons in `Cr^(3+) (g)` isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `Cr-4s^(2) 3d^(5), Cr^(+)-4s^(0) 3d^(3)` The correct answer is 3.
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| 1032. |
The spectrum of `He` is expected to be similar to.A. HB. `Li^(+)`C. NaD. `He^(+)` |
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Answer» Correct Answer - B (b) Both `He` and `Li^+` contain `2` electrons each. |
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| 1033. |
Uncertainty in position is twice the uncertainty in momentum. Uncertainty in velocity is :A. `sqrt((h)/(pi))`B. `(1)/(2m) sqrt((h)/(pi))`C. `(1)/(2m) sqrt(h)`D. `(h)/(4 pi)` |
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Answer» Correct Answer - C ( c) `Delta x = 2 Delta p` `Delta x. Delta p = (ħ)/(2) = (h)/(4 pi)` `rArr 2 Delta p. Delta p = (ħ)/(2)` `2(m Delta V)^2 = (ħ)/(2) , (Delta V)^2 = (ħ)/(4 m^2)` `rArr5 Delta V = (sqrt(ħ))/(2 m)`. |
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| 1034. |
Radiation of `lambda`=155 nm was irradiated on Li(work function =5 eV) plante . The stopping potential (in eV) is.A. 4 eVB. 5 eVC. 3 eVD. 1 eV |
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Answer» Correct Answer - C `V=(hc)/lambda_(2)-phi_(0)=3 eV` |
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| 1035. |
`A` hydrogen-like atom has ground state binding energy `122.4eV`. Then :A. its atomic number is `3`B. an electron of `90 eV` can excite it to a higher stateC. an `80 eV` electron cannot excite it to a higher stateD. an electron of `8.2 eV` and a photon of `91.8 eV` are emitted when a `100 eV` electron interacts with it |
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Answer» Correct Answer - A::C::D `Delta x = 2Deltap" "Deltax .2Deltap = (h)/(4pi)` `2(Deltap)^(2)=(h)/(4pi)" "(Deltav)^(2)=(h)/(8pim^(2))` `Deltav=(1)/(2m)sqrt((h)/(2pi))" "Deltav=(1)/(2m)sqrt(h)` |
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| 1036. |
The order of filling of electrons in the orbitals of an atom will beA. 3d, 4s, 4p, 4d, 5sB. 4s, 3d, 4p, 5s, 4dC. 5s, 4p, 3d, 4d, 5sD. 3d,4p,4s,4d,5s |
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Answer» Correct Answer - B As per Aufbau principle |
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| 1037. |
An orbital with `l = 0` isA. Symmetrical about X axis onlyB. Symmetrical about Y axis onlyC. Spherically symmetrqacalD. Unsymetrical |
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Answer» Correct Answer - C `l = 0` or s orbital |
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| 1038. |
For a given principal level `n = 4` the energy of its subshells is of the odrerA. `s lt d lt f lt p`B. `s lt p lt d lt f`C. `d lt f lt p lt s`D. `s lt p lt f lt d` |
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Answer» Correct Answer - B `s lt p lt d lt f` |
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| 1039. |
Statement : The energy radiated per unit volume , i.e., energy density in block body radiation depends upon the temperature . Explanation : Green light in never emitted in black body radiations .A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - D Both are facts |
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| 1040. |
F-orbitals in presence of magnetic field are :A. Non-degeneratgB. Five fold-degeneratC. Sevenfole-degeneratD. None of the above |
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Answer» Correct Answer - A In presence of magnetic field oritals are non-degenceratie i.e, possess different energy levels . |
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| 1041. |
how fast is an elecreon moving if it has a wavelength equal to the distance travelled in one second ?A. ` sqrt m/h`B. `sqrt h/m`C. `sqrt h/p`D. `sqrt h/(mKE)` |
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Answer» Correct Answer - B `lambda= h/(mu)` if `lambda =u` ` u = sqrt h/m` . |
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| 1042. |
The ratio of radius of first orbit in hydrogen to the radius of first orbit in deuterium will be :A. `1:1`B. `1:2`C. `2:1`D. `4:1` |
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Answer» Correct Answer - C |
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| 1043. |
Statement : All s-orbitla in H-atom corresponds to a non-zero probability density at nucleus . Explanation : The probability density is given by ` psi^2` and ` psi prop e ^(Z2//2a_0)`.A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - C ` psi prop e^(-Zr//2a_0)` at `r=0, psi^2 prop e^(-0) prop 1`. |
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| 1044. |
Given wave function represents which orbital of hydrogen ``Psi=1/4 1/(sqrt2pi)(1/(alpha_(0)))^(5//2) re^(-r//2alpha)cos 0` (where 0=angle from z-axis)A. `2P_(y)`B. `2P_(z)`C. `3P_(y)`D. `3P_(Z)` |
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Answer» Correct Answer - B |
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| 1045. |
Which orbital is represented by the complete wave funcion ` psi _(420)` ?A. `4d`B. `3d`C. `4p`D. `4s` |
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Answer» Correct Answer - A For 4d-orbital ` n=4, l=2 m=0` |
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| 1046. |
In hydrogen atom, energy of first excited state is `- 3.4 eV`. Then, `KE` of the same orbit of hydrogen atom is.A. `+ 3.4 eV`B. `+6.8 eV`C. `-13.6 eV`D. `+13.6 eV` |
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Answer» Correct Answer - A (a) `:. ` Total energy `(E_(n))=KE +PE` In first excited state = `(1)/(2) mv^(2)+[-(Ze^(2))/(r)]` `=+(1)/(2) (Ze^(2))/(r)-(Ze^(2))/(r)` Energy of first excited state is 3.4 eV `-3.4 eV=-(1)/(2) (Ze^(2))/(r)` `:. KE=(1)/(2)(Ze^(2))/(2)` `=+3.4 eV` |
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| 1047. |
The total energy of the electron in the hydrogen atom in the ground state is `- 13.6 eV`. The `KE` of this electron is.A. `13.6 eV`B. zeroC. `- 13.6 eV`D. `6.8 eV` |
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Answer» Correct Answer - A (a) `because KE = (1)/(2) mv^2 = (Ze^2)/(2 r) , TE = - (Ze^2)/(2 r)`. |
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| 1048. |
Which of the following is electronic configuration of `Cu^(2+) (Z = 29)` ?A. `[Ar]4s^(1)3d^(8)`B. `[Ar]4s^(2)3d^(10)4p^(1)`C. `[Ar]4s^(1)3d^(10)`D. `[Ar]3d^(9)` |
| Answer» Correct Answer - D | |
| 1049. |
If value of azimuthal quantum number `l` is `2`, then total possible values of magnetic quantum number will be.A. 7B. 5C. 3D. 2 |
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Answer» Correct Answer - B (b) If value of `1` is `2` then `m = -2, -1, 0, + 1, + 2.m = -1 + 1` including zero. (`5` values of magnetic quantum number). |
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| 1050. |
The electronic transition that emits maximum energy is [n= represents orbit]A. `n_(5) rarr n_(4)`B. `n_(4) rarr n_(3)`C. `n_(2) rarr n_(1)`D. `n_(3)rarr n_(2)` |
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Answer» Correct Answer - C The energy difference b/n first and second orbits is more. |
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