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,
$\frac{1242\text{eV nm}}{100\text{nm}}=12\cdot 42\text{eV}$
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
${\lambda }_1=\frac{1242\text{eV}\text{nm}}{10\cdot 2\text{eV}}=122\text{nm}$

${\lambda }_2=\frac{1242\text{eV}\text{nm}}{12\cdot 1\text{eV}}=103\text{nm}$
These wavelengths will have low intensity in the transmitted beam.