Question

On de-excitation of an electron from higher energy orbit to second orbit energy of electron decreases by 7.6 eV in a hydrogen like ion. What will be the approx de-Broglie wavelength of electron in the original exicted state? [Given : I.E. of hydrogen= 13.6 eV]

• A. 5 $\stackrel{o}{A}$
• B. 2 $\stackrel{o}{A}$
• C. 9 $\stackrel{o}{A}$
• D. 3 $\stackrel{o}{A}$

Answer 1

The correct option is C 9 $\stackrel{o}{A}$

$\Delta E_{excelation}=-13.6z^2(\frac{1}{n{f}^2}-\frac{1}{n_1^2})$

$-7.6=13.6[\frac{1}{4}-\frac{1}{n^2}]$

$\frac{1}{n^2}=\frac{-1}{4}+\frac{76}{136}$

$=\frac{440}{554}$

If in $n^{th}$ ergy orbital

$\frac{440}{544}\times 13.6=11=\frac{hC}{-1}=E_A=\frac{-{13.62}^{2}eV}{n^2}$

$\lambda =9A^o$