Question

Suppose potential energy between electron and proton at separation $r$ is given by $U=k\ln r$, where $k$ is constant. For such hypothetical hydrogen atom, the ratio of energy difference between energy levels of ($n=1$ to $n=2$) and ($n=2$ to $n=4$) is

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $3$

Answer 1

The correct option is A $1$

The electrostatic force can be given in terms of potential energy as,
$F=-\frac{dU}{dr}=-\frac{k}{r}$

As we know that, this force provides for the centripetal force required for electron to move in circles to the proton. Hence we can write:
$\frac{k}{r}=\frac{m{v}^2}{r}\dots \dots \left(i\right)$

Energy level in an orbit is given by:
$E_n=\frac{1}{2}m{v}^2+k\ln r\dots \dots \left(ii\right)$

Angular momentum is given by:
$mvr=\frac{nh}{2\pi }\dots \dots \left(iii\right)$

Solving $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get

$E_n=\frac{k}{2}(1+\ln \left(\frac{n^2{h}^2}{4{\pi }^{2}mk}\right))$

Ratio of energy difference:
$\frac{E_2-E_1}{E_4-E_2}=\frac{\ln \left(\frac{4h^2}{4{\pi }^{2}mk}\right)-\ln \left(\frac{h^2}{4{\pi }^{2}mk}\right)}{\ln \left(\frac{16h^2}{4{\pi }^{2}mk}\right)-\ln \left(\frac{4h^2}{4{\pi }^{2}mk}\right)}$

$\frac{E_2-E_1}{E_4-E_2}=\frac{\ln \left(4\right)-\ln \left(1\right)}{\ln \left(16\right)-\ln \left(4\right)}=\frac{1}{1}$