Question

Monochromatic radiation of wavelength $\lambda $ is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. Find the value of $\lambda $ .

• A. $102nm$
• B. $100nm$
• C. $97.5nm$
• D. $94.5nm$

Answer 1

Answer: (C)

As the hydrogen atoms emit radiation of six different wavelengths, some of them must have been excited to $\text{n} = 4$ , The energy in $\text{n} = 4$ state is
$\text{E}_{4} = \frac{\text{E}_{1}}{4^{2}} = - \frac{1 3 \cdot 6 \text{e} \text{V}}{1 6} = - 0 \cdot 8 5 \text{eV.}$
The energy needed to take a hydrogen atom from its ground state to $\text{n} = 4$ is
$\text{13} \text{.6 eV - 0} \text{.85 eV = 12} \text{.75 eV}$ .
The photons of the incident radiation should have $12.75 \, \, \text{eV}$ of energy. So
$\frac{\text{hc}}{\lambda } = 1 2 \cdot 7 5 \text{eV}$
or , $\lambda = \frac{\text{hc}}{1 2 \cdot 7 5 \text{eV}}$
$= \frac{1 2 4 2 \text{eV} \text{nm}}{1 2 \cdot 7 5 \text{eV}}$
$\text{= 97} \text{.5 nm}$ .