Question

A hydrogen atom in the ground state is excited by radiations of wavelength $975A^{\circ }$. The energy state to which the atom is excited, is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

The energy of the incident photon is:
$E=\frac{hc}{\lambda }$

$E=\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-19}}eV$

$E=\frac{19.875\times {10}^{-26}}{1560\times {10}^{-29}}eV$

$E=0.01274\times {10}^{3}eV=12.74eV$

Due to absorption of this energy the electron in ground state gets excited to (say)$n^{th}$ state. Hence,
$E=-\frac{Rch}{n^2}-(-\frac{Rch}{1^2})$
$E=10.96\times {10}^6\times 3\times {10}^8\times 6.6\times {10}^{-34}[1-\frac{1}{n^2}]$
$12.74=13.6[1-\frac{1}{n^2}]$
$1-\frac{1}{n^2}=\frac{12.74}{13.6}$
$\frac{1}{n^2}=1-0.9367$
$\frac{1}{n^2}=0.0633$
$n^2=15.79\approx 16$
$n=4$