Q. The electric potential between a proton and an electron is given by $V = V_{0} \text{ln} \frac{r}{r_{0}}$ where r₀ is a constant. Assuming Bohr’s model to be applicable, write variation of $r_{n}$ with $n$ , $n$ being the principal quantum number.

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A

$r_{n} \propto n$

B

$r_{n} ∝ \frac{1}{n}$

C

$r_{n} ∝ n^{2}$

D

$r_{n} ∝ \frac{1}{n^{2}}$

Solution:

U = eV = eV₀ ln $\left(\right. \frac{r}{r_{0}} \left.\right)$
$| F | = | - \frac{\text{d} U}{\text{d} r} | = \frac{e V_{0}}{r}$
This force provides necessary centripetal force.
Hence, $\frac{m v^{2}}{r} = \frac{e V_{0}}{r}$
Or $v = \sqrt{\frac{e V_{0}}{m}}$ ... (i)
Moreover, $m v r = \frac{n h}{2 \pi }$ ... (ii)
Dividing Eq. (ii) by (i), we have
$\textit{mr}=\left(\frac{\textit{nh}}{2 \pi }\right)\sqrt{\frac{\textit{m}}{\left(\text{eV}\right)_{0}}}$ or $\text{r}_{\text{n}} \propto \text{n}$

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Q. The electric potential between a proton and an electron is given by V=V0lnrr0V = V_{0} \text{ln} \frac{r}{r_{0}} where r0 is a constant. Assuming Bohr’s model to be applicable, write variation of rnr_{n} with nn , nn being the principal quantum number.

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1. A thin semicircular conducting ring of the radius RR is falling with its plane vertical in a horizontal magnetic induction →B\overset{ \rightarrow }{B} . At the position MNQ, the speed of the ring is vv and the potential difference developed across the ring is

2. An ideal gas at 27∘C27 \, ^\circ C is compressed adiabatically to 8/278/27 of its original volume. If γ=5/3\gamma =5/3 , then the rise in temperature is

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