Question

In a hydrogen like atom, an electron is orbiting in an orbit having quantum number '$n$'. Its frequency of revolution is found to be $13.2\times {10}^{15}Hz$. Energy required to remove this electron from the orbit is $54.4eV$. In the time of 7 nanosecond the electron jumps from$\text{'}n\text{'}th$ orbit to the orbit having quantum number $\frac{n}{2}$'. If $\tau$ be the average torque acted on the electron during the above process, then find $\frac{T\times {10}^{27}}{3}$ (in $Nm$). (Given:$\frac{h}{\pi }=2.1\times {10}^{-34}Js,$ frequency of revolution of electron in the ground state of H atom is ${\nu }_0=6.6\times {10}^{15}Hz$ and ionization energy of H atom is $E_0=13.6eV$)

Answer 1

$\nu =v_0\frac{Z^2}{n^3}\Rightarrow \frac{Z^2}{n^3}=2.$
And, $E=E_0\frac{Z^2}{n^2}\Rightarrow \frac{Z^2}{n^2}=4.$
Solving above two equation, we get $n=2,Z=4$.
Now,
$L=mvr=\frac{nh}{2\pi }\&\Delta L=\tau \Delta t=\Delta n\frac{h}{2\pi }$
Now,

$\Delta n=n-\frac{n}{2}=\frac{n}{2}=1$
And,
$\Rightarrow \tau =\frac{\Delta n}{\Delta t}\times \frac{h}{2\pi }=\frac{2.1\times {10}^{-34}}{7\times {10}^{-9}\times 2}\Rightarrow \tau =15\times {10}^{-27}\therefore \frac{T\times {10}^{27}}{3}=5$