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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.
| 801. |
The number of electrons present in each `alpha`-particle isA. 2B. 3C. 0D. 1 |
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Answer» Correct Answer - C `alpha`-particles are di-positive helium ions. `therefore Z=2` A=4 Number fo electrons=0 Helium atom contains two electrons. Its loses the two electrons to form `He^(+2)` ion. |
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| 802. |
How could Bohr explain the stability of an atom ? |
| Answer» Energy of electron in an orbit. | |
| 803. |
The photons of light having a wavelength `4000 Å` are necessary to provide `1.00 J` of energy are.A. `6.023 xx 10^23`B. `6.023 xx 10^18`C. `2.01 xx 10^18`D. `2.01 xx 10^23` |
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Answer» Correct Answer - C ( c) Energy of one photon =`hv = h. (c)/(lamda) = ((6.62 xx 10^-34)(3.0 xx 10^8))/(4000 xx 10^-10) = 4.965 xx 10^-19 J` Number of photons `= (1.00)/(4.965 xx 10^-19) = 2.01 xx 10^18`. |
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| 804. |
Write the electronic arrangment for the following elements. Also give the geometrical representations. (a) Oxgyen (b) Argon (c) Calcium (d) Potassium |
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Answer» Conclusions : (i) Atom to be empty (ii) Nucleus |
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| 805. |
The atomic transition gives rise to the radiation of frequency `(10^(4)HMz)`. The change in energy per mole of atoms taking place wiould be:A. `3.99xx10^(-6)J`B. `3.99J`C. `6.62xx10^(-24)J`D. `6.62xx10^(-30)J` |
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Answer» Correct Answer - B `E=Nhv` `=6.023xx10^(23)xx6.626xx10^(-34)xx10^(4)xx10^(6)=3.99J` |
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| 806. |
The photons of light having a wavelength `4000 Å` are necessary to provide `1.00 J` of energy are. |
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Answer» Energy of one photon `=hv=h.(c)/lamda` `=((6.62xx10^(-34))(3.0xx10^(8)))/(4000xx10^(-10))` `=4.965xx10^(-19)J` Number of photons `=(1.00)/(4.965xx10^(-19))=2.01xx10^(18)`. |
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| 807. |
The number of photons of light having wavelength 100 nm which can provide 1 J energy is nearly:A. `10^(7)` photonsB. `5xx10^(18)` photonsC. `5xx10^(17)` photonsD. `5xx10^(7)` photons |
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Answer» Correct Answer - C `E=(nhc)/(lamda)` `n=(Elamda)/(hc)=(1xx100xx10^(-9))/(6.626xx10^(-34)xx3xx10^(8))=5xx10^(17)`. |
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| 808. |
A metal surface having ` v_0` as threshode frequency is incident by light of frequency (v) then select the correct :A. ` u = sqrt(2 h(lambda_0 - lambda))`B. ` u = sqrt(2 h(lambda_0 - lambda))`C. ` u = sqrt(2 h(lambda_0 - lambda)/(m))`D. ` u = sqrt (2 h(lambda_0 - lambda)/(m))` |
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Answer» Correct Answer - A::B::D `hv = hv_0 + 1/2 mu^2` ` :. U = sqrt(2h (v-v_0)/m)0 = sqrt((2hc (lambda_0 - lambda)/(mx lambda_0 xx lambda))` ` hv = w + 1/2 mu^2` ` :. U = sqrt((2 (hv-w))/m)`. |
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| 809. |
What is the number of proton of light with wavelength `300`nm that provide `2J` of energy ? |
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Answer» Let N proton of light provide `2J` energy so `:. E =Nhv= (Nhc)/(lambda)` or `N = (E lambda)/(hc)` `= (2 xx 300 xx 10^(-9)m)/(6.262 xx 10^(-34) xx 3 xx 10^(6) ms^(-1))` `= 3.01 xx 10^(18)` photons |
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| 810. |
The approximate radius of a H-atom is `0.05` nm, and that of proton is `1.5 xx 10^-15` m. Assuming both hydrogen atom and the proton to be spherical, calculate fraction of the space in an atom of hydrogen that is occupied by the nucleus.A. `2.7 xx 10^-14`B. `2.7 xx 10^-10`C. `5 xx 10^-14`D. `2 xx 10^-16` |
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Answer» Correct Answer - A (a) `(V_("nucleus"))/(V_("atom")) = ((4)/(3) pi r_1^3)/((4)/(3) pi r_2^3) = ((1.5 xx 10^-15)/(0.5 xx 10^-10))^3 = 2.7 xx 10^-14`. |
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| 811. |
What is the ratio of mass of an electron to the mass of a proton ?A. `1 : 2`B. `1 : 1`C. `1 : 1837`D. `1 : 3` |
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Answer» Correct Answer - C ( c) Electron `(1)/(1837)` time lighter than proton so their mass ratio will be `1 : 1837` ratio `= (1.67 xx 10^-27)/(9.1 xx 10^-31) = 1837`. |
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| 812. |
The element having no neutron in the nucleus of its atom isA. hydrogenB. nitrogenC. heliumD. boron |
| Answer» Correct Answer - A | |
| 813. |
Nuclear particles responsible for holding all nuclear composition of an atom would lead to a change in:A. electronsB. neutronsC. positronsD. mesons |
| Answer» Correct Answer - D | |
| 814. |
Which fo the following statement (s) is are correct ?A. The electronic configruation of ` cr` is `[Ar] 3d^5 m 4s^1`B. The magnetic quatum no. may have abegative valueC. In silve atom ` 23` eelctrons have a spin of one type and ` 24` of the oppositive type ( an. of . Ag is 74)D. The oxidation no. of nitrogen in ` N_3 H ` is ` -3` |
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Answer» Correct Answer - A::B::C `._(24)Cr: 1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^5 4s^1` (d-acquires half -filled notrue ) `m=- l` to `+ l` include zero In silver `._(47)Ag: 1s^2 , 2s^2 3p^6 , 3s^2 3p^6 3d^(10)` `4s^2, 4p^6 4d^(10) , 5s^1` Thus ` 24` electron have , ` s = + 1//2` while rest ` 23 ` have ` s=- 1//2` . |
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| 815. |
Decrease in atomic mnumber is boserved during :A. alpha-emissionB. `beta-emissionC. positron-emissionD. electron capture |
| Answer» Correct Answer - A::C::D | |
| 816. |
The introduction of a neutron into the nuclear composition of an atom would lead to a change in:A. Its atomic massB. Its atomic numberC. the chemical nature of the atomD. number of the electron also |
| Answer» Correct Answer - A | |
| 817. |
Calculate the frequency `&` energy of a photon of wave length `4000A`. |
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Answer» `(a)` Calculation of frequency : ` lambda=4000Å` `lambda=4000xx10^(-10)m` `because " "v=(C )/(lambda)` `therefore " "v=(3xx10^(8)m//sec)/(4xx10^(-7)m)=0.75xx10^(15)sec^-1` `=7.5 xx10^(14)sec^(-1)` `(b)` Calculation of energy : `E=hv=6.626xx10^(-34)"Joule"xx7.5xx10^(14)sec^(-1)` `=4.96xx10^(-19)`Joule |
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| 818. |
The number of electrons in the nucleus of `C^(12)` isA. 6B. 12C. 0D. 3 |
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Answer» Correct Answer - C ( c) In the nucleus of an atom only proton and neutrons are present. |
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| 819. |
An atom which has lost one electron would beA. Negatively chargedB. Positively chargedC. Electrically neutralD. Carry double positive charge |
| Answer» Correct Answer - B | |
| 820. |
Number of neutron in `C^12` isA. 6B. 7C. 8D. 9 |
| Answer» Correct Answer - A | |
| 821. |
The ratio of energies of photons with wavelength `2000A^(0)` and `4000A^(0)` isA. `1//4`B. 4C. `1//2`D. 2 |
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Answer» Correct Answer - D `(E_(1))/(E_(2)) = (lambda_(2))/(lambda_(1))` |
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| 822. |
`CO` has same electrons as of the ion that is isoelectronic with `CO` is.A. `N_2^+`B. `CN^-`C. `O_2^+`D. `O_2^-` |
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Answer» Correct Answer - B (b) `CO` and `CO^-` are isoelectronic. `CO = 6 + 8 = 14` and `CN^(-)= 6 + 7 + 1 = 14`. |
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| 823. |
The increasing order of the ionic radii of the given isoelectronic species is :-A. `K^(+),S^(2),Ca^(2+),Cl^(-)`B. `Cl,Ca^(2+),K^(+),S^(2)`C. `S^(2-),Cl^(-),Ca^(2+),K^(+)`D. `Ca^(2+),K^(+),Cl,S^(2-)` |
| Answer» Correct Answer - D | |
| 824. |
Among `._(10)A^(20) ._(11)B^(21) ._(11)C^(22)` and `._(122)D^(22)` the isobar combination isA. A & BB. B & CC. C & DD. A & D |
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Answer» Correct Answer - C isobars have same mass number |
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| 825. |
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) sqrth`D. `sqrt((h)/(4 pi))` |
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Answer» Correct Answer - C |
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| 826. |
The energy of electron is maximum atA. NucleusB. Ground stateC. First excitedD. Infinite distance from the nucleus |
| Answer» Correct Answer - D | |
| 827. |
Calculate the angular frequency of an electron occupying the second Bohr orbit of `He^(o+)` ion |
| Answer» `2. 067 xx 10^(16) sec^(-1)`, | |
| 828. |
Which of the following best explains light both as a stream of particles and wave motion ?A. DiffractionB. `lamda=h//p`C. InterferenceD. Photoelectric effect |
| Answer» Correct Answer - B | |
| 829. |
Number of nomal planes ( planes of zero electron densit ) in the `d_(xy)` orbital isA. 1B. 2C. 0D. 4 |
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Answer» Correct Answer - 2 `xz` and `yz` are planes with zero electron density for orbital. |
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| 830. |
An electron is accelerated to one -tenth the velocity of light suppose its velocity can be measurted with a precision of `1%` what must be the uncertain with qa precition of `1 %` what must be the uncertainty in position ? |
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Answer» Correct Answer - A::C::D `Delta = [(1)/(100) xx (1)/(10)c] = ( 3 xx 10^(8) m s^(-1))/(1000) = 3 xx 10^(5) m s^(-1)` `:. Delta p Delta x ge (h)/(4pi)` `Delta x ge (1)/(Delta p)xx(h)/(4pi)` `ge (1)/(m Delta v) ,(h)/(4pi)` `ge (1)/(9.1 xx 10^(-31)(kg)) xx (1)/( 3 xx 10^(5)m s^(-1)) xx (6.626 xx 10^(-34)(Js))/(4 xx 3.14)` `ge 1.93 xx 10^(-10)m` |
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| 831. |
Which is correct for any kind of species?A. `(E_(2)-E_(1))gt (E_(3)-E_(2)) gt (E_(4)-E_(3))`B. `(E_(2)-E_(1)) gt (E_(3)-E_(2)) lt (E_(4)-E_(3))`C. `(E_(2)-E_(1))=(E_(3)-E_(2))=(E_(4)-E_(3))`D. `(E_(2)-E_(1))=1//4(E_(3)-E_(2))=1//9(E_(4)-E_(3))` |
| Answer» Correct Answer - A | |
| 832. |
What is uncerainty in location of photod of wavelenght 5000Å if wavelenght is known to an accuracy of 1 pm?A. `7.96xx10^(-14)`B. `0.02m`C. `3.9xx10^(-8)m`D. None |
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Answer» Correct Answer - B `Deltalambda.Deltax ge(lambda^(2))/(4pi)` `Deltax ge2xx10^(2)m` |
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| 833. |
Number of visible lines when an electron returns from 5th orbit to ground state in H spectrum :A. 5B. 4C. 3D. 10 |
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Answer» Correct Answer - C |
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| 834. |
From hydrogen gas discharge tube, light is emitted. Emitted light is used to emit electron from sodium metal of work function `phi= 1.82 eV` the transition in hydrogen atom takes place from:A. `n_(1)=4"to"n_(2)=1`B. `n_(1)=4"to"n-(2)=3`C. `n_(1)=3"to"n_(2) =1`D. `n_(1)=4"to"n_(2)=2` |
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Answer» Correct Answer - d |
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| 835. |
Which of the following electromagnetic radiation possesses highest amount of energy ?A. X- raysB. Gamma raysC. Radio wavesD. Cosmis rays |
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Answer» Correct Answer - D |
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| 836. |
For which of the following particles will it be most difficult to experimentally verify the de Broglie relationship ?A. A dust particleB. An electronC. A protonD. An `alpha`-particle projected closer to the center of the atom will experience a lesser deviation as compared to the particle projected away from the center. |
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Answer» Correct Answer - A |
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| 837. |
What is the ratio of volume of atom of the volume of nucleus?A. `10^(10)`B. `10^(12)`C. `10^(15)`D. `10^(20)` |
| Answer» Correct Answer - C | |
| 838. |
Electromagnetic radiation with maximum wavelengths is :A. ultravioletB. radiowavesC. X-raysD. infrared |
| Answer» Correct Answer - B | |
| 839. |
Calculate the energy in kilocalorie per mol of the photons of an electromagnetic radiation of wavelength 7600Ã…. |
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Answer» `lamda=6000` Ã…`=7600xx10^(-8)cm` `c=3xx10^(10)cms^(-1)` Frequency, `v=(c)/(lamda)=(3xx10^(10))/(7600xx10^(-8))=3.947xx10^(14)s^(-1)` Energy of one photon `=hv=6.61xx10^(-12)xx6.02xx10^(23)` `=15.71xx10^(11)erg` Energy of one mole of photos in kilocalorie. `=(15.71xx10^(11))/(4.185xx10^(10))" "[1kcal=4.185xx10^(10)erg]` `=37.538` kcal per mol. |
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| 840. |
If the wavelength of photon emitted from an electron jump `n=4` to `n=2` in a `H-` like species is `1216 Å` then the species is `:`A. `H-` atomB. `He^(+)` ionC. `Li^(2+)`ionD. `Be^(3+)`ion |
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Answer» Correct Answer - 2 Energy of photon `=(124000)/(1216)=10.2eV=(H)_(2+1)=(He^(+))_(4rarr2)` so, species is `He^(+)` ion |
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| 841. |
The ionization potential of hydrogen atom is 13.6 eV. The wavelength of the energy radiation required for the ionization of H-atomA. 1911 nmB. 912 nmC. 68 nmD. 91.2nm |
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Answer» Correct Answer - D `Delta E = (hc)/(lambda)` |
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| 842. |
In a hydrogen atom, if energy atom, if energy of an electron in group state is `- 13.6 eV`, then that in the `2^(nd)` excited state is :A. ` -1 . 5 1 eV`B. ` -3 4eV`C. ` -6 . 04 eV`D. `-6.04eV` |
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Answer» Correct Answer - A 2bd excited state mens third energy level . ` E_3 = E_1 /3_2 =- ( 13.6)/9 =- 1. 51eV` . |
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| 843. |
The fraction of volume occupied by the nucleus with respect to the total volume of an atom is.A. `10^(-15)`B. `10^(-5)`C. `10^(-30)`D. `10^(-10)` |
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Answer» Correct Answer - A |
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| 844. |
The quantum number that was proposed to explain the Zeeman effect isA. mB. lC. sD. n |
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Answer» Correct Answer - A Magnetic quantum number |
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| 845. |
Choose the correct statement(s) regarding the photo-electric effect.A. No electrons are ejected, regardless of the intensity of the radiation, unless the frequency exceeds a thershold value characteristic of the metalB. The kinetic energy of the ejected electrons varies linearly with the frequency of the incident radiation and its intensityC. Even at low intensities, electrons and ejected immediately if the frequency is above the threshold valueD. An intense and a weak beam of monochromatic radiations differ in having number of photons and not in the energy of photons |
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Answer» Correct Answer - A::C::D Photo electric effect concept |
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| 846. |
The fraction of volume occupied by the nucleus with respect to the total volume of an atom is.A. `10^-15`B. `10^-5`C. `10^-30`D. `10^-10` |
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Answer» Correct Answer - A (a) Volume fraction `("Volume of nucleus")/("Total vol. of atom")` =` ((4//3)pi(10^-13)^3)/((4//3)pi(10^-8)^(3)) = 10^-15`. |
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| 847. |
`{:(Column I,Column II),((A)"Radius of" n^(th) "orbit",(p)"Inversely proportional to Z"),((B)"Energy of" n^(th)"orbit",(q)"Integral multiple of"(h)/(2pi)),((C)"Velocity of electron" 10^(-10)m "in the" n^(th) "orbit",(r)"Proportional to"),((D)"Angular momentum",(s)"Inversely proportional to n"),(,(t)"Inversely proportional too" n^(2)):}` |
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Answer» Correct Answer - A-p,r;B-r,t; C-s; D-q `r_(n,z) alpha (n^(2))/(2) E_(n,z) alpha-(z^(2))/(n^(2)), V_(n,z) alpha(z)/(n) mvr alphan` |
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| 848. |
Select the incrorect statement (s) :A. A negative enrgy state is more stable relative to a zero erngy state .B. The kinetic erngy fo photoelectrons emitted by a photosensitive surface depends on the intensity of radiation .C. In a photoelectric emission process , the maximum energy of the photoelectron increases with increasing intensity of the incident light .D. The value for ` psi_(1 s)^2` increase with the distance of the electron form the nucleus . |
| Answer» Correct Answer - B::C::D | |
| 849. |
The radius of an atomic nucleus is of the order ofA. `10^(-10) `cmB. `10^(-13) `cmC. `10^(-15) `cmD. `10^(-8) `cm |
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Answer» Correct Answer - B The radius of nucleus is of the order of `1.5xx10^(-13)` to `6.5xx10^(-13)` cm or `1.5` to 6.5 Fermi (1Fermi =`10^(-13)` cm). |
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| 850. |
Select the correct statements (s) :A. The co-ordinate `(theta)` in ` psi (r, theta , phi)` is the angle form the positive z-axis ( the nroth pole ) representing lattiude .B. The co-ordinate `(theta)` in ` psi ( r, theta , phi)` is the angle about the z-axis represnting the longitude ,C. Tehe schrodinger equation restricts the elcron in a H-atom decrease exponentially towres zereo or ( r) increase .D. The co-ordinate `(theta)` in ` psi ( r, theta , phi)` is the angle about the z-axis represnting the longitude , |
| Answer» Correct Answer - A::B::D | |