Question

An electron in the third energy level of an exited He${}^{\oplus }$ ion returns back to the ground state. The photon emitted in the process is absorbed by a stationary hydrogen atom in the ground state. The velocity of the photo electron ejected from the hydrogen atom in m s${}^{-1}$ :

• A. 3.49 $\times$ 10${}^6$ ms${}^{-1}$
• B. 4.49 $\times$ 10${}^6$ ms${}^{-1}$
• C. 5.59 $\times$ 10${}^6$ ms${}^{-1}$
• D. 8.59 $\times$ 10${}^6$ ms${}^{-1}$

Answer 1

The correct option is A 3.49 $\times$ 10${}^6$ ms${}^{-1}$

Given,
He${}^{\oplus }$(Z = 2);n = 3$\to$ n = 1
$\Delta$E = 13.6(2)^2[$\frac{1}{1^2}$ - $\frac{1}{3^2}$] = (13.6 $\times$ 4 $\times$ $\frac{8}{9}$)
Also as we know,
($\Delta$E - $\varphi$) = KE = $\frac{1}{2}$mV${}_e^2$
IE of H atom
$\Rightarrow$ (13.6 $\times$ 4 $\times$ $\frac{8}{9}$ - 13.6) = $\frac{1}{2}$mV${}_e^2$
$\Rightarrow$ ($\frac{23}{9}\times 13.6)$ $\times$ $\frac{2}{9.1\times {10}^{-31}}$ = V${}_e^2$
$\Rightarrow$ V${}_e$ = $\sqrt{1.222\times {10}^{13}}$ = 3.49 $\times$ 10${}^6$ ms${}^{-1}$