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QJEEAdv20Phe04e: JEE Advanced 2020 Physics - Class 12 JEEAdv20Phe04e Solution: A particle of mass m moves in circular orbits with potential…

A particle of mass m moves in circular orbits with potential energy V(r) = Fr, where F is a positive constant and r is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle’s orbit is denoted by R and its speed and energy are denoted by v and E, respectively, then for the nth orbit (here h is the Planck’s constant)

Multiple Choice Options

  • A R∝n13R \propto {n^{{1 \over 3}}} and v∝n23v \propto {n^{{2 \over 3}}}
  • B R∝n23R \propto {n^{{2 \over 3}}} and v∝n13v \propto {n^{{1 \over 3}}}
  • C E=32(n2h2F24π2m)13E = {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{{1 \over 3}}}
  • D E=2(n2h2F24π2m)13E = 2{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{{1 \over 3}}}

Step-by-Step Verified Solution

∣Force∣=∣−dVdr∣⇒F=mv2r|Force| = \left| { - {{dV} \over {dr}}} \right| \Rightarrow F = {{m{v^2}} \over r} .....(i)

mvr=nh2πmvr = {{nh} \over {2\pi }} .....(ii)

From Eqs. (i) and (ii), we get

∴\therefore F=mr(nh2πmr)2F = {m \over r}{\left( {{{nh} \over {2\pi mr}}} \right)^2} ∵\because (v=nh2πmr)\left( {v = {{nh} \over {2\pi mr}}} \right)

or F=mn2h24π2m2r3F = {{m{n^2}{h^2}} \over {4{\pi ^2}{m^2}{r^3}}}

or F=n2h24π2mr3F = {{{n^2}{h^2}} \over {4{\pi ^2}m{r^3}}}

r3∝ n2{r^3} \propto \,{n^2} (as F = constant)

r∝ n2/3r \propto \,{n^{2/3}}

v=nh2πmrv = {{nh} \over {2\pi mr}}

v∝nn2/3v \propto {n \over {{n^{2/3}}}}

or v∝n1/3v \propto {n^{1/3}}

∴\therefore KE=12mv2=Fr2KE = {1 \over 2}m{v^2} = {{Fr} \over 2}

∵\because TE=PE+KE=Fr+Fr2TE = PE + KE = Fr + {{Fr} \over 2}

⇒3Fr2=TE\Rightarrow {{3Fr} \over 2} = TE

or TE=E=3F2×(n2h24π2rnF)1/3TE = E = {{3F} \over 2} \times {\left( {{{{n^2}{h^2}} \over {4{\pi ^2}rnF}}} \right)^{1/3}}

=32(n2h2F24π2m)1/3= {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{1/3}}

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