Question

In a hydrogen like atom, an electron is revolving in an orbit having quantum number $n$. Its frequency of revolution is found to be $1.32\times {10}^{15}\text{Hz}.$ Energy required to completely remove this electron to infinity from the above orbit of atom is $54.4\text{eV}$. In a time interval of $7$ nanoseconds the electron jumps back to orbits having quantum number $n/2.$ If the average torque acting on the electron during the above process is $\tau \text{Nm},$ then find the value of $\tau \times {10}^{27}$ (given $\frac{h}{\pi }=2.1\times {10}^{-34}\text{J-s},$) frequency of revolution of electron in the ground state of H-atom $=6.6\times {10}^{15}\text{Hz}$_______

Answer 1

Frequency of electron revolving in $n$th orbit, $\nu ={\nu }_0\frac{Z^2}{n^3}....\left(1\right)$

Energy of electron in $n$th orbit, $E=E_0\frac{Z^2}{n^2}....\left(2\right)$

From the data given in the question,

$\left(1\right)$ gives, $\frac{Z^2}{n^3}=2....\left(i\right)$

and $\left(2\right)$ gives, $\frac{Z^2}{n^2}=4.....\left(ii\right)$

$\text{Solving (i) and (ii),}n=2andZ=4$

We know that, $L=mvr=\frac{nh}{2\pi }$

Also, $\Delta L=\tau \Delta t=\Delta n\left(\frac{h}{2\pi }\right)$

$\Rightarrow \tau =\frac{\Delta n}{\Delta t}\times \frac{h}{2\pi }=\frac{2-1}{7\times {10}^{-9}}\times 1.05\times {10}^{-34}=0.15\times {10}^{-25}\text{Nm}\text{or}15\times {10}^{-27}\text{Nm}$