Question

A sample of atomic hydrogen gas is irradiated by a beam of ultraviolet radiation having a wavelength between $\text{100 nm}$ and $\text{200 nm}$ . Assuming that the atoms are in the ground state, the wavelength which will have low intensity in the transmitted beam is

• A. $\text{104 nm}$
• B. $\text{103 nm}$
• C. $\text{105 nm}$
• D. $\text{100 nm}$

Answer 1

Answer: (B)

The energy of a photon corresponding to $\lambda =100nm$ is,
$\frac{1 2 4 2 \text{eV nm}}{1 0 0 \text{nm}}=12\cdot 42\text{eV}$
Similarly, the corresponding to $\lambda =200nm$ is $6.21eV$ .
The energy needed to take the atom from the ground state to the first excited state is
$E_{2}-E_{1}=13.6eV-3.4eV=10.2eV$
The energy needed to take the atom from the ground state to the second excited state is
$E_{3}-E_{1}=13.6eV-1.5eV=12.1eV$
The energy needed to take the atom from the ground state to the third excited state is
$E_{4}-E_{1}=13.6eV-0.85eV=12.75eV$ , and so on.
Thus, $10.2eV$ photons and $12.1eV$ photons have a large probability of being absorbed from the given range of $6.21eV$ to $12.42eV$ .
The corresponding wavelengths are
$\lambda _{1} = \frac{1 2 4 2 \text{eV} \text{nm}}{1 0 \cdot 2 \text{eV}} = 1 2 2 \text{nm}$

$\lambda _{2} = \frac{1 2 4 2 \text{eV} \text{nm}}{1 2 \cdot 1 \text{eV}} = 1 0 3 \text{nm}$
These wavelengths will have low intensity in the transmitted beam.