Question

If a hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength $975\mathring{A}$, the number of different lines possible in the resulting spectrum and the longest wavelength among them are

• A. $6,18270\mathring{A}$
• B. $4,18720\mathring{A}$
• C. $4,304\mathring{A}$
• D. $6,8208\mathring{A}$

Answer 1

The correct option is B $4,18720\mathring{A}$

Formulae:

$E=\frac{hc}{\lambda }$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-9}}eV$

$=12.75eV$

$12.75=13.6(\frac{1}{1^2}-\frac{1}{n^2})$

$\frac{1}{n^2}=1-\frac{12.75}{13.6}$

$=\frac{0.85}{13.6}$

$=\frac{1}{16}$

$\therefore n=4$

$\lambda =\frac{hc}{E_{4\to 3}}$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{13.6(\frac{1}{3^2}-\frac{1}{4^2})\times 1.6\times {10}^{-9}}m$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{13.6\times 7\times 1.6\times {10}^{-9}}m$

$\lambda =1.88\times {10}^{-6}m$

$\Rightarrow \lambda =18800\dot{A}$