Question

A photon of wavelength 12.4 nm is used to emit an electron from ground state of He${}^{+}$. Calculate de-broglie wavelength of emitted electron

• A. 1.2$\dot{A}$
• B. 2.2$\dot{A}$
• C. 1.8$\dot{A}$
• D. 2.7$\dot{A}$

Answer 1

Answer: (C)

1.8$\dot{A}$

$E=-13.6\frac{z^2}{n^2}=-13.6\times \frac{2^2}{12}=-13.6\times 4$
$=-54.4eV$
Energy of $12.4nm=\frac{1240}{12.4}eV$ $E=\frac{12400}{124A^o}eV$
$=100eV$
$\therefore$ Energy left $=100eV-54.4eV$
$=45.6eV$
de-Broglie wavelength
$\lambda =\frac{1.227}{\sqrt{v}}nm\frac{1.227}{\sqrt{45.6}}nm$
$=0.18nm=1.8A^o$