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Question: The electric potential between a proton and an electron is given by\(V = V_{0}\ln\frac{r}{r_{0}}\), ...

The electric potential between a proton and an electron is given byV=V0lnrr0V = V_{0}\ln\frac{r}{r_{0}}, where r0r_{0} is a constant. Assuming Bohr’s model to be applicable, write variation of rnr_{n} with n, n being the principal quantum number

A

rnnr_{n} \propto n

B

rn1/nr_{n} \propto 1/n

C

rnn2r_{n} \propto n^{2}

D

rn1/n2r_{n} \propto 1/n^{2}

Answer

rnnr_{n} \propto n

Explanation

Solution

Potential energy U=eV=eV0lnrr0U = eV = eV_{0}\ln\frac{r}{r_{0}}

∴ Force F=dUdr=eV0rF = - \left| \frac{dU}{dr} \right| = \frac{eV_{0}}{r}.

The force will provide the necessary centripetal force.

Hence mv2r=eV0r\frac{mv^{2}}{r} = \frac{eV_{0}}{r}v=eV0mv = \sqrt{\frac{eV_{0}}{m}}…..(i) and

mvr=nh2πmvr = \frac{nh}{2\pi} …..(ii)

Dividing equation (ii) by (i) we have mr=(nh2π)meV0mr = \left( \frac{nh}{2\pi} \right)\sqrt{\frac{m}{eV_{0}}}

or rn