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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.
| 1701. |
Uncertainty principle gave the concept ofA. probabilityB. An orbitalC. Physical meaning of `Psi` the `Psi^2`D. All the above |
| Answer» Correct Answer - D | |
| 1702. |
On the basis of heisenhergs uncertainty principle show that the electron correct exist within the nucleus |
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Answer» The ratio of the nucleus is of order of `10^(-13) cm ` and thus the uncertainty in position of the electron (i.e `Delta x)` d it is within the nucless will be `10^(-13) cm ` `Now Delta x- Delta u ge (h)/(4pi m)` `:. Delta u = (6.626 xx 10^(-27))/(4 xx 3.14 xx 9.108 xx 10^(-28) xx 10^(-15))` ` = 5.79 xx 10^(12)cm s^(-1)` i.e. the order of velocity of the electron will be `100 `times greater than the velocity of light .White is impossible .Thus possibility of the electron to exist in the nucless is zero |
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| 1703. |
Which one of the following configuration represents a noble gas ?A. `1s^2 , 2s^2 2p^6 , 3s^2`B. `1s^2 , 2s^2 2p^6 , 3s^1`C. `1s^2 , 2p^2 2p^6`D. `1s^2 , 2s^2 2p^6 , 3s^2 3p^6 , 4s^2` |
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Answer» Correct Answer - C `1s^2, 2s^2, 2p^6` represents a noble gas electronic configuration. |
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| 1704. |
The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number `m_(s)` = - 1 /2, is |
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Answer» Correct Answer - 9 So, electrons with spin quantum numbe =`-1/2` will be 1+3+5=9 |
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| 1705. |
The atomic masses of He and Ne are 4 and 20 amu respectively . The value of the de Broglie wavelength of He gas at`-73.^(@)C` is 'M' times that of the de Broglie wavelength of Ne at `727.^(@)C.` M is |
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Answer» Correct Answer - 5 `lambda=h/sqrt(2m(KE)) " " KE prop T` `lambda_(He)/lambda_(Ne)=sqrt((m_(Ne)KE_(Ne))/(m_(He)KE_(He)))=sqrt((20xx1000)/(4xx200))=5` |
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| 1706. |
When the canal ray experiment was conducted by taking helium gas in the discharge tube, e/m value of the particles was found to be less under low voltage and it was found to be more under high voltage . How do you explain this ? |
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Answer» (i) The amount of energy required to cause ionisation . (ii) Factors affecting e/m values (iii) Charge on helium gas under low pressure and high pressure. (iv) The effect of charge on e/m. |
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| 1707. |
Whne canal rays experiment is conducted with hydrogen gas, scientists were found to give particles with different `(e)/(m)` values. Justify. |
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Answer» (i) e/m depends on number of protons and neutrons (ii) existence of isotopes (iii) variation in e/m for isotopes. |
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| 1708. |
Arrange the following statements given by various scientists in chronological order :A. 4 3 12B. 42 31C. 2431D. 4321 |
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Answer» (i) Atoms of the same elements are identical in all respects. (ii) Assumpation of thinly spread positively-charged mass. (iii) Calculation of the diameters of the nucleus and the atom (iv) Calculation of energy and radius of orbit |
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| 1709. |
A trinegative ion of an element has 8 electrons in its M shell. The atomic number of the element isA. 15B. 18C. 20D. 16 |
| Answer» Since electronic configuration of the triengative ion is 2,8,8 the electronic configuration of the neutral atom is 2,8,5 and its atomic number is 15. | |
| 1710. |
Some of the `alpha`-rays deflect in acute and obtuse angles due to the presence of the ________in the centre of the atom. |
| Answer» positive charge | |
| 1711. |
An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV, find the de-Broglie wavelength of the electron.A. `sqrt((150)/(3.4)) "Å"`B. `sqrt((150)/(6.8))"Å"`C. `sqrt((150)/(3.4))` nmD. `sqrt((150)/(6.8))` nm |
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Answer» Correct Answer - A |
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| 1712. |
Calculate the unceertainty in position of an electron if the uncertainty in its velocity is `5.7 xx 10^(5) m s^(-1), h = 6.6 xx 10^(-24) kg m^(2) s^(-1)` mass of electron `= 9.1 xx 10^(-13 kg) ` |
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Answer» Correct Answer - A::D `1 xx 10^(10)m` Hint: `Delta x - Delta p = (h)/(4pi) ` or` Delta x - Delta v = (h)/(4pi) ` `Delta x = (h)/(4pi m Delta v) ` `m_(s) = 9.1 xx 10^(-27) g Delta V = 5.7 xx 10^(3) ms^(-1)` |
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| 1713. |
The maximum energy is present in any electron atA. nucleusB. ground stateC. first excited stateD. infinite distance from the nucleus |
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Answer» Correct Answer - D |
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| 1714. |
An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV, find the de-Broglie wavelength of the electron.A. `66.5 Å`B. `6.66Å`C. `60.6Å`D. `6.06Å` |
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Answer» Correct Answer - B `n = 2 2pi r_(n) = n lambda_(n) rArr lambda_(n) = (2pir_(n))/(n)` |
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| 1715. |
Calculate the unceertainty in the momentum of a particle if the uncertainity in its position is ` 6.6 xx 10^(-32) m ` |
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Answer» Correct Answer - A::D `7.97 xx 10^(-4) kg m s^(-1)` Hint: `Delta x - Delta p = (h)/(4pi) ` ` :. Delta p = (h)/( Delta s 4pi)` |
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| 1716. |
The angular momentum of electron in an excited H atom is `(h)/(pi)` . The P.E. of electron will be :A. 6.8 eVB. 3.4 eVC. `-6.8 eV `D. `3.4 eV` |
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Answer» Correct Answer - C |
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| 1717. |
The work functions `(W_(0))` of K, Na, Li, Mg and Cu are 2.25,2.30,2.42,3.70 and 4.80 eV respectively. How many of these metals do not undergo photoelectric effect when a radiation of wavelength 450 nm is allowed to fall on them? `(1eV=1.602xx10^(-19)J)`A. 2B. 1C. 3D. 5 |
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Answer» Correct Answer - A Ejection of photoelectron takes place when the absorbed quantum energy exceeds threshold energy. `E_("absorbed")=(hc)/(lamda)` `=(6.626xx10^(-34)xx3xx10^(8))/(450xx10^(-19))` `=4.417xx10^(-19)J` `(4.417xx10^(-19))/(1.602xx10^(-19))eV` =2.75eV Thus, Mg and Cu i.e., two metals will not eject photoelectron from the given radiation. |
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| 1718. |
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n = 4` to `n = 2` of `He^(o+)` spectrum ?A. `n_(1) = 1` "to" `n_(2) = 2`B. `n_(1) = 2 "to" n_(2) = 4`C. `n_(1) = 1 "to" n_(2) = 3`D. `n_(1) = 2 "to" n_(2) = 3` |
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Answer» Correct Answer - A `barv _(H_2^o+) = (1)/(lambda_(H_2^o+)) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`.....(i) `barv _(He^o+) = (1)/(lambda_(H^o+)) = RZ^(2)((1)/(2^(2)) - (1)/(4^(2)))` `= R xx 4 ((1)/(4) - (1)/(16))` `=- R xx ((4)/(4)- (4)/(16)) = R xx (1 - (1)/(4))`.....(ii) Compairing equation (i) and (ii) `:. (1)/(n_(1)^(2)) = ,n_(1) = 1` `(1)/(n_(2)^(2)) = (1)/(4) , n_(2) = 2` |
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| 1719. |
Calculate the wavelength and energy for radiation emitted for the electron transition from infinite `(oo)` to stationary state of the hydrogen atom `R = 1.0967 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34) J s ` and `c = 2.979 xx 10^(8) m s^(-1)` |
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Answer» Correct Answer - A::B::C::D `n_(1) = 1,n_(2) = oo,(1)/(lambda) = 1.09678 xx 10^(7) [(1)/(1^(2)) - (1)/(oo^(2))]` `lambda = 9.11 xx 10^(6) m ` `lambda = 9.11 xx 10^(-6) m` `E = hv = b(c )/(lambda) = Rhc (since (1)/(lambda) = R)` ` = 1.09678 xx 10^(7) xx 6.62 xx 10^(-34) xx 3 xx 10^(8) m` `= 217.9 xx 10^(-20) J` `217.9 xx 10^(-23) kJ` |
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| 1720. |
Calculate the wavelength and energy of radiation emitted for the electron transition from infinite `(oo)` to first stationary state of the hydrogen atom. `R = 1.0967 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34) J s ` and `c = 2.979 xx 10^(8) m s^(-1)` |
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Answer» `(1)/(lamda)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `n_(1)=1 and n_(2)=infty` `(1)/(lamda)=R[(1)/(1^(2))-(1)/((infty)^(2))]=R` or `lamda=(1)/(R)=(1)/(1.09678xx10^(7))=9.11xx10^(-8)m` We know that, `E=hv=h.(c)/(lamda)=6.6256xx10^(-34)xx(2.9979xx10^(8))/(9.11xx10^(-8))` `=2.17xx10^(-18)J`. |
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| 1721. |
A certain dye absorbs `4530A^(@)` and fluoresence at `5080A^(@)` these being wavelength of maximum absorption that under given condition `47%` of the absorbed energy is emitted. Calculate the ratio of the no of quanta emitted to the number absorbed. |
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Answer» Correct Answer - `0.527` E of light absorbed in one photon `=(hc)/(lambda_("absorbed"))` Let `n_(1)` photons are absorbed , therefore, Total energy absorbed `=(n_(1)hc)/(lambda_("absorbed"))` Now, E of light re-emitted out in one photon `=(hc)/(lambda_("emitted"))` Let `n_(2)` photons are re-emitted then Total energy re-emitted out `=n_(2)xx(hc)/(lambda_(emitted))` As given `E_("absorbed")xx(47)/(100)=E_("re-emitted out")` `(hc)/(lambda_("absorbed"))xxn_(1)xx(47)/(100)=n_(2)xx(hc)/(lambda_("emitted"))` `therefore (n_(1))/(n_(2))=(47)/(100)xx(lambda_("emitted"))/(lambda_("absorbed"))=(47)/(100)xx(5080)/(4530)` `therefore (n_(1))/(n_(2))=0.527` |
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| 1722. |
The specific charge of photon is `9.6xx10^7 C kg^(-1)` then for an `alpha`-particle it will beA. `38.4xx10^7 " C kg"^(-1)`B. `19.2xx10^7 " C kg"^(-1)`C. `2.4xx10^7 " C kg"^(-1)`D. `4.8xx10^7 " C kg"^(-1)` |
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Answer» Correct Answer - D `(g/m)_alpha=1/2(q/m)_p=1/2xx9.6xx10^7 =4.8xx10^7 C kg^(-1)` |
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| 1723. |
`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 state(c). an `80 eV` electron cannot excite it to a higher state(d). an electron of `8.2 eV` and a photon of `91.8 eV` are emitted when a `100 eV` electron interacts with itA. Its atomic number is 3B. a photon of 90 eV can excite it to a higher stateC. a 80 eV photon cannot excite it to a higher stateD. to excite an `e^(-)` minimum 91.8 eV photon is required |
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Answer» Correct Answer - A::C::D |
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| 1724. |
For a ball of mass 198.6 g to be located within `0.1 "Å"` , what will be the minimum uncertainity in velocity ?A. `2.6 xx 10^(-17)` cm/sB. `2.6 xx 10^(-18)` m/sC. `2.6 xx 10^(-21)` cm/sD. `2.6 xx 10^(-21)` m/s |
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Answer» Correct Answer - C |
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| 1725. |
The constancy of e/m ratio for electron shows thatA. Electrons mass is `(1)/(1837)th` of mass of protonB. Electrons are universal particles of all metterC. Electrons are produced in discharge tube onlyD. None of these |
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Answer» Correct Answer - B electrons are universal particles |
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| 1726. |
An `alpha`- particle has initial kinetic energy of 25 eV and it is accelerated through a potential difference of 150 volt . If a photon has initial kinetic energy of 25 eV and it is accelerated through a potential difference of 25 volt , then find the approximate ratio of the final wavelengths associated with the proton and the `alpha`-particle .A. 5B. 4C. 3D. 2 |
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Answer» Correct Answer - A |
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| 1727. |
What hydrogen-like ion has the wavelength difference the first lines of Balmer and Lyman series equal to `59.3 nm`? `R_(H) = 109.678 cm^(-1)` |
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Answer» We have `(1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`For first line of Balmer series `(1)/(lambda_(B)) =R_(H).Z^(2) [(1)/(2^(2)) - (1)/(3^(2))] = (5)/(36) xx R_(H) xx Z^(2)` or `lambda_(B) = (36)/(5.R_(H).Z^(2))`.......(i) For first line of Lyman series `(1)/(lambda_(L)) =R_(H).Z^(2) [(1)/(1^(2)) - (1)/(2^(2))] = (3)/(4) xx R_(H) xx Z^(2)` or `lambda_(L) = (4)/(3R_(H).Z^(2))`.....(ii) Given `lambda_(B) - lambda_(L) = 59.3 xx 10^(-7) cm ` or `(36)/(5R_(H).Z^(2)) -(4)/(3R_(H).Z^(2))= 59.8 xx 10^(-7)` `(1)/(5R_(H).Z^(2))[7.2 - 1.333] = 59.3 xx 10^(-7)` or `Z^(2) = (5.867)/(R_(H) xx 59.3 xx 10^(-7))` `= (5.867)/(109678 xx 59.8 xx 10^(-7))` `:. Z = 3` `:. "H-like atom is" Li^(2+)` |
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| 1728. |
If `lambda_(1)` and `lambda_(2)` denote the de-Broglie wavelength of two particles with same masses but charges in the ratio of `1:2` after they are accelerated from rest through the same potential difference, thenA. `lambda_(1) = lambda_(2)`B. `lambda_(1) lt lambda_(2)`C. `lambda_(1) gt lambda_(2)`D. `lambda_(1) lt sqrt(lambda_(2))` |
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Answer» Correct Answer - C `lambda = (h)/(sqrt(2.m.q.v))` |
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| 1729. |
Assertion (A) : In an atom, the velocity of electron in the higher orbits keeps on decreasing Reason (R ) : Velocity of electron in inversely proportional to the radius of the orbitA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
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Answer» Correct Answer - A Due to same electronic configuration. |
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| 1730. |
Calculate the wavelength a particle of mass `m = 6.6 xx 10^(-27) kg` moving with kinetic energy `7.425 xx 10^(-13) J ( h = 6.6 xx 10^(-34) kg m^(2) s ^(-1)` |
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Answer» Correct Answer - A::B::D `6.67 xx 10^(-15)m` Hint :`lambda = (h)/(P)= (h)/(mv)` `KE = (1)/(2) mv^(2) = 7.425 xx 10^(-13) J` `V = (( 2 xx 7.425 xx 10^(-13))/(6.6 xx 10^(-30)))^(1//2)` |
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| 1731. |
Which of the following statement is correct in relation to the hydrogen atom :A. 3s-orbital is lower in energy than 3p-orbitalB. 3p-orbital is lower in energy than 3d-orbitalC. 3s-and 3p-orbital are of lower energy than 3-orbital.D. 3s,3p-and 3d-orbitals all hae the same energy. |
| Answer» Correct Answer - D | |
| 1732. |
The number of `d` electrons in `Ni` (at.no `= 28`) is equal to that of theA. s and p-electrons in `F^(-)`B. p-electrons in Ar (At. No=18)C. d-electron in `Ni^(2+)`D. total number of electron in N (At No.=7) |
| Answer» Correct Answer - C | |
| 1733. |
The number of radial nods of 4p-orbital is:A. 4B. 3C. 2D. 1 |
| Answer» Correct Answer - C | |
| 1734. |
Which amog the followng statements is/are correct?A. `Phi^(2)` represents the atomic orbitalsB. The number of peaks in radial distribution is `(n-l)`C. Radial probability density `rho_(nl)(r)=4pir^(2)R_(nl)^(2)(r)`D. A node is point in space where the wave function `(Phi)` has zero amplitude. |
| Answer» Correct Answer - A::B::C::D | |
| 1735. |
Select the correct configurations among the following:A. `Cr(Z=24):[Ar]3d^(5),4s^(1)`B. `Cu(Z=29):[Ar]3d^(10),4s^(1)`C. `Pd(Z=46):[Kr]4d^(10),5s^(0)`D. `Pt(Z=78):{Xe]4d^(10)4s^(2)` |
| Answer» Correct Answer - A::B::C | |
| 1736. |
Which among the following are correct about angular momentum of electron?A. `2h`B. `1.5(h)/(pi)`C. `2.5h`D. `0.5(h)/(pi)` |
| Answer» Correct Answer - A::B::D | |
| 1737. |
Which element is represented by the following electronic configuration ? .A. NitrogenB. FluorineC. OxygenD. None of these |
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Answer» Correct Answer - D (d) `._10 Ne = 1 s^2 2 s^2 2 p^6`. |
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| 1738. |
Calculate the radius of `2^(nd)` excited state of `Li^(+2)` |
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Answer» `2^(nd)` excited state, means `e^(-)` is present in `3^(rd)` shell so. `r_(3)" "= 0.529xx(3xx3)/(3)=0.529xx3Å = 1.587 Å` |
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| 1739. |
The orbital cylindrically symmetrical abut x-axis is :A. `P_z`B. `P_y`C. `P_x`D. `d_(xz` |
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Answer» Correct Answer - C `p_(x)` orbial has rwo lobes on x-axis. |
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| 1740. |
Calculate the energy of `Li^(+2)` atom for `2^(nd)` excited state. |
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Answer» `E =-13.6xx(Z^(2))/(n^(2))` `because Z=3` and `e^(-)` exist in `2^(nd)` excited state, means `e^(-)` present in `3^(rd)` shell i.e. `n=3` `therefore E=-13.6 xx((3)^(2))/((3)^(2))=-13.6 eV//"atom"` |
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| 1741. |
In sample of hydrogen atoms in ground state, electron make transition from ground state to particular excited state, where path length is four times of debroglie wavelenght . Electrons make back transition to the ground state prouducing all possible photons. Ve,cocity of electron in fianal excited dtate in the above sample of hydrogen atom isA. `(2.18xx10^(6))/4m//s`B. `(2.18xx10^(6))/8m//s`C. `(2.18xx10^(6))/2m//s`D. None of these |
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Answer» Correct Answer - A `2pir_(n)=nlambda=4lambda rArrn=4` `V=2.18xx10^(6)(z)/(n)m//s=(2.18xx10^(6))/(4)m//s` |
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| 1742. |
In sample of hydrogen atoms in ground state, electron make transition from ground state to particular excited state, where path length is four times of debroglie wavelenght . Electrons make back transition to the ground state prouducing all possible photons. Debroglie wavelenght of electron which is emitted from the fianl excited dtate of above hydrogen sample, when photon of 1eV collide with itA. `5sqrt2Å`B. 82.6 nmC. 1240 nmD. `10sqrt10 Å` |
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Answer» Correct Answer - D Final exicted state, `n =4, E_(4)=-0.85 eV` K.E. of the electron after collding with photon of 1 eV`=(1-0.85) eV=0.15 eV` `lambda=sqrt((150)/(0.15))=10sqrt(10)` |
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| 1743. |
The vapours of Hg absord some electron accelerated by a potiential diff. of 4.5 volt as a result of which light is emitted . If the full energy of single incident `e^(-)` is supposed to be converted into light emitted by single Hg atom , find the wave no. of the light |
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Answer» Correct Answer - `3.63xx10^(6)m^(-1)` `4.5 eV = (1240)/(lambda(nm))" "(1)/(lambda)=(4.5)/(1240)` `(1)/(lambda)=0.0036nm^(-1)" "1nmrarr10^(-9)m^(-1)` `0.0036nm^(-1)" "1nmrarr10^(-9)m^(-1)` |
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| 1744. |
Which of the following is not permissible arrangement of electrons in an atom ?A. n=4, l=0, m=0, s=-1/2B. n=5, l=3, m=0, s=+1/2C. n=3, l=2, m=-3, s=-1/2D. n=3, l=2, m=-2, s=-1/2 |
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Answer» Correct Answer - C (c) If n=3, l=0 to (3-1)=0, 1, 2 m=-l to +l = -2, -1, 0 +1, +2 s=`+-(1)/(2)` Therefore, option (c) is not a permissible set of quantum numbers. |
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| 1745. |
If `n = 6`, the correct sequence for filling of electrons will be.A. `nstonpto(n-1)dto(n-2)f`B. `nsto(n-2)fto(n-1)dtonp`C. `nsto(n-1)dto(n-2)ftonp`D. `nsto(n-2)ftonpto(n-1)d` |
| Answer» Correct Answer - B | |
| 1746. |
if n=6, the correct sequence of filling of electrons will beA. `ns to np (n-1) d to (n-2) f`B. `ns to n (n-2) f to (n-1) d to np`C. `ns to (n-1) d to (n-2) f to np`D. `ns to (n-2) f to np to (n-1) d` |
| Answer» Correct Answer - B | |
| 1747. |
If `n = 6`, the correct sequence for filling of electrons will be.A. ns`rarr` (n-1)d`rarr`(n-2)f`rarr`npB. ns `rarr`(n-2)f`rarr`np`rarr`(n-1)dC. ns`rarr`np`rarr`(n-1)d`rarr`(n-2)fD. ns`rarr`(n-2)f`rarr`(n-1)d`rarr`np |
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Answer» Correct Answer - D (d) `6s rarr 4f rarr 5d rarr 6p ` for n=6 |
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| 1748. |
The energy of an electron in `n^"th"` orbit of hydrogen atom isA. `13.6/n^4 eV`B. `13.6/n^3 eV`C. `13.6/n^2 eV`D. `13.6/n` eV |
| Answer» Correct Answer - C | |
| 1749. |
For balmer series in the spectrum of atomic hydrogen the wave number of each line is given by `bar(V) = R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` where `R_(H)` is a constant and `n_(1)` and `n_(2)` are integers. Which of the following statements (s), is (are correct) 1. As wave length decreases the lines in the series converge 2. The integer `n_(1)` is equal to 2. 3. The ionisation energy of hydrogen can be calculated from the wave numbers of three lines. 4. The line of shortest wavelength corresponds to = 3.A. 1, 2 and 3B. 2 , 3 and 4C. 1,2 and 4D. 2 and 4 only |
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Answer» Correct Answer - C (1)Beyond a certain wavelength the line spectrum becomes band spectrum. (2)For Balmer series `n_1=2` (4)The line of longest wavelength in this series , `n_2=3` |
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| 1750. |
In a certain electronic transition in the hydrogen atoms from an initial state `(1)` to a final state `(2)`, the difference in the orbit radius `((r_(1)-r_(2))` is 24 times the first Bohr radius. Identify the transition-A. `4 to 1`B. `4 to 2`C. `4 to 3 `D. `3 to 1` |
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Answer» Correct Answer - C |
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