Question

The ionisation energy of the hydrogen atom is given to be 13.6 eV. A photon falls on a hydrogen atom which is initially in the ground state and excites it to the (n=4) state.
The wavelength of the photon is:

• A. $973.5\stackrel{o}{A}$
• B. $210\stackrel{o}{A}$
• C. $389.3\stackrel{o}{A}$
• D. $849.6\stackrel{o}{A}$

Answer 1

The correct option is A $973.5\stackrel{o}{A}$

The transition is $n=1\to n=4$
The expression for the wavelength is $\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Substitute values in the above expression.
$\frac{1}{\lambda }=109678c{m}^{-1}[\frac{1}{1^2]}-\frac{1}{4^2}]$
Hence, the wavelength is $973.5A^0$.