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 move this electron from the atom to the above orbit is $54.4eV$. In a time of $7$nano second the electron jumps back to orbit having quantum number $n/2$. If $\tau$ be the average torque acted on the electron during the above process, then $\tau =(10+x)\times {10}^{27}$ in $Nm$. Find the value of $x$.

Given: $h/\lambda =2.1\times {10}^{-34}J-s$, frequency of revolution of electron in the ground state of $H$ atom $v_0=6.6\times {10}^{15}$ and ionization energy of $H$ atom $E_0=13.6eV$

Answer 1

Let the atomic number of the atom be $Z$.

The energy of $nth$ orbit is given as $54.4eV$

$\therefore 54.4eV=13.6\times \frac{Z^2}{n^2}eV⟹Z=2n$

Also the frequency of revolution of the electron in $nth$ orbit, ${\nu }_n=6.6\times {10}^{15}\frac{Z^2}{n^3}$

$\therefore 13.2\times {10}^{15}=6.6\times {10}^{15}\frac{4n^2}{n^3}⟹n=2$

Time taken by the electron to jump from $nth$ orbit to $\frac{n}{2}th$ orbit, $\Delta t=7ns$

The change in angular momentum of the electron due to the transition,

$\Delta L=\frac{nh}{2\pi }-\frac{\left(\frac{n}{2}\right)h}{2\pi }=\frac{nh}{4\pi }$

$\therefore$ Average torque acted, $\tau =\frac{\Delta L}{\Delta t}=\frac{2\times 6.63\times {10}^{-34}}{4\times 3.14\times 7\times {10}^{-9}}=15\times {10}^{-27}Nm$