Question

The difference between the angular momentum of two orbits of $H{e}^{+2}$ is $\frac{2h}{\pi }$ . The energy of an electron present in the higher orbit is -1.51 eV. Indentify the lower orbit.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

The energy of electron $E=\frac{-13.6\times Z^2}{n^2}ev$
$\therefore 1.51=\frac{-13.6\times 4}{n^2}$
$[\because Z=2]$
$\therefore n=6$
The difference in angular momentum of two orbits is
$(mvr{)}_x-(mvr{)}_y=\frac{n_{x}h}{2\pi }-\frac{n_{y}h}{2\pi }$
$\therefore \frac{2h}{\pi }=(n_x-n_y)\frac{h}{2\pi }$
$n_x-n_y=4$
$\therefore$ The other orbit $\left(n_y\right)$ is $2^{nd}$ orbit.