Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Atoms and Nuclei question

2023 · 6 Apr · Shift 2 · Q60
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Atoms and Nuclei
  5. /2023 · 6 Apr · Shift 2 · Q60

Atoms and Nuclei question

2023 · 6 Apr · Shift 2 · Q60

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
A small particle of mass mmm moves in such a way that its potential energy U=12m ω2r2U=\frac{1}{2} m ~\omega^{2} r^{2}U=21​m ω2r2 where ω\omegaω is constant and rrr is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of nth n^{\text {th }}nth  orbit will be proportional to,
  1. A
    n\sqrt{n}n​
  2. B
    n2n^{2}n2
  3. C
    1n\frac{1}{n}n1​
  4. D
    nnn
View written solutionFree

Correct answer: A

  1. Given potential energy

The particle moves in a central potential

U(r)=12mω2r2.U(r)=\frac{1}{2}m\omega^2 r^2.U(r)=21​mω2r2.

For a circular orbit, the required centripetal force is provided by the central force due to this potential.

  1. Find the force from the potential

The radial force is

F(r)=−dUdr=−mω2r.F(r)=-\frac{dU}{dr}=-m\omega^2 r.F(r)=−drdU​=−mω2r.

So its magnitude is

∣F∣=mω2r.|F|=m\omega^2 r.∣F∣=mω2r.
  1. Use circular motion condition

For circular motion of radius rrr and speed vvv,

mv2r=mω2r.\frac{mv^2}{r}=m\omega^2 r.rmv2​=mω2r.

Cancelling mmm,

v2r=ω2r⇒v2=ω2r2.\frac{v^2}{r}=\omega^2 r \quad\Rightarrow\quad v^2=\omega^2 r^2.rv2​=ω2r⇒v2=ω2r2.

Hence,

v=ωr.v=\omega r.v=ωr.
  1. Apply Bohr quantization of angular momentum

Bohr's quantization condition is

mvr=nℏ.mvr=n\hbar.mvr=nℏ.

Substitute v=ωrv=\omega rv=ωr:

m(ωr)r=nℏ.m(\omega r)r=n\hbar.m(ωr)r=nℏ.

So,

mωr2=nℏ.m\omega r^2=n\hbar.mωr2=nℏ.

Thus,

r2=nℏmω.r^2=\frac{n\hbar}{m\omega}.r2=mωnℏ​.

Therefore,

rn=nℏmω.r_n=\sqrt{\frac{n\hbar}{m\omega}}.rn​=mωnℏ​​.
  1. Proportionality

So the radius of the nthn^{\text{th}}nth orbit is proportional to

rn∝n.r_n\propto \sqrt{n}.rn​∝n​.
  1. Check options
  • A: n\sqrt{n}n​ ✅
  • B: n2n^2n2 ❌
  • C: 1n\frac{1}{n}n1​ ❌
  • D: nnn ❌

Hence the correct option is A.

PreviousNext

More from Atoms and Nuclei

  • Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is x9​×10−10 m. The…2023 · Numerical
  • For a nucleus AA​X having mass number A and atomic number Z A. The surface energy per nucleon (bs​)=−a1​A2/3. B. The Coulomb contribution to the…2023 · MCQ
  • A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in…2023 · Numerical
  • The waves emitted when a metal target is bombarded with high energy electrons are2023 · MCQ
  • The ratio of wavelength of spectral lines Hα​ and Hβ​ in the Balmer series is 20x​. The value of x is ​.2023 · Numerical
  • The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be2023 · MCQ
  • If 917 Ao​ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ​Ao​.2023 · Numerical
  • The energy of He+ ion in its first excited state is, (The ground state energy for the Hydrogen atom is −13.6 eV) :2023 · MCQ