Question

A hydrogen atom at rest is in ground state. It is struck by a $H{e}^{\prime }$ ion in first excited state. Assuming the collision to be head on and the mass of $H{e}^{+}$ to be four times that of hydrogen atom, find the least value of kinetic energy of incoming particle which can excite both the particles to second excited state.

Answer 1

We know that, ${ϵ}_n=-13.6\times \frac{z^2}{n^2}$

Ground state energy of $H=-13.6\times \frac{1^2}{1^2}=-13.6eV$

second excited state energy of $H=-13.6\times \frac{1^2}{3^2}$

$=\frac{-13.6}{9}=-1.52eV$

Now,

First excited state of $H{e}^{+}=-13.6\times \frac{2^2}{2^2}$

$=-13.6eV$

second excited state of Net $H{e}^{+}=-13.6\times \frac{2^2}{3^2}eV=-13.6\times \frac{4}{9}eV=-6.04eV$

Now, Let KE be the minimum kinetic energy of $H{e}^{+}$ ion, So that both single electron species get excited, then

(Ground state energy of $H$)+(First excited state energy of $H{e}^{+}$)$+KE$=(2nd excited state energy of $H$)+(2nd excited state energy of $H{e}^{+}$)

$(-13.6)+(-13.6)+KE=(-1.52)+(-6.04)eV$

$-27.2+KE=-7.56$

$KE=-7.56+27.20$

$KE=+19.64eV$