Question

A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.

Answer 1

Given:
Wavelength of light, λ = 100 nm = 100 $\times$ 10^$-9$ m
Energy of the incident light$\left(E\right)$ is given by
$E=\frac{hc}{\mathrm{\lambda }}$
Here,
h = Planck's constant
λ = Wavelength of light
$$\begin{gathered} \therefore E=\frac{1242}{100} \\ E=12.42\mathrm{eV} \end{gathered}$$

(a)
Let E₁​ and E₂ be the energies of the 1st and the 2nd state, respectively.
Let the transition take place from E₁ to E₂.
Energy absorbed during this transition is calculated as follows:
Here,
n_​​​1 = 1
n_​​​2 = 2
Energy absorbed (E') is given by
$$\begin{gathered} E\text{'}=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right) \\ =13.6\left(\frac{1}{1}-\frac{1}{4}\right) \\ =13.6\times \frac{3}{4}=10.2\mathrm{eV} \end{gathered}$$
Energy left = 12.42 eV − 10.2 eV = 2.22 eV
​Energy of the photon = $\frac{hc}{\lambda }$
Equating the energy left with that of the photon, we get
$$\begin{gathered} 2.22\mathrm{eV}=\frac{hc}{\mathrm{\lambda }} \\ 2.22\mathrm{eV}=\frac{1242}{\mathrm{\lambda }} \end{gathered}$$
or λ = 559.45 = 560 nm
Let E₃ be the energy of the 3rd state.
Energy absorbed for the transition from E₁ to E₃ is given by
$$\begin{gathered} E\text{'}=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right) \\ =13.6\left(\frac{1}{1}-\frac{1}{9}\right) \\ =13.6\times \frac{8}{9}=12.1\mathrm{eV} \end{gathered}$$
Energy absorbed in the transition from E₁ to E₃ = 12.1 eV (Same as solved above)

Energy left = 12.42 − 12.1 = 0.32 eV

$$\begin{gathered} 0.32=\frac{hc}{\mathrm{\lambda }}=\frac{1242}{\mathrm{\lambda }} \\ \mathrm{\lambda }=\frac{1242}{0.32} \\ =3881.2=3881\mathrm{nm} \end{gathered}$$
Let E₄ be the energy of the 4th state.
Energy absorbed in the transition from E₃ to E₄ is given by
$$\begin{gathered} E\text{'}=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right) \\ =13.6\left(\frac{1}{9}-\frac{1}{16}\right) \\ =13.6\times \frac{7}{144}=0.65\mathrm{eV} \end{gathered}$$
Energy absorbed for the transition from n = 3 to n = 4 is 0.65 eV
Energy left = 12.42 − 0.65 = 11.77 eV
Equating this energy with the energy of the photon, we get
$$\begin{gathered} 11.77=\frac{hc}{\mathrm{\lambda }} \\ \mathrm{or}\mathrm{\lambda }=\frac{1242}{11.77}=105.52 \end{gathered}$$
The wavelengths observed in the transmitted beam are 105 nm, 560 nm and 3881 nm.

(b)
If the energy absorbed by the 'H' atom is radiated perpendicularly, then the wavelengths of the radiations detected are calculated in the following way:
$$\begin{gathered} E=10.2\mathrm{eV} \\ \Rightarrow 10.2=\frac{hc}{\lambda } \\ \mathrm{or}\lambda =\frac{1242}{10.2}=121.76\mathrm{nm}\approx 121\mathrm{nm} \\ E=12.1\mathrm{eV} \\ \Rightarrow 12.1=\frac{hc}{\lambda } \\ \mathrm{or}\lambda =\frac{1242}{12.1}=102.64\mathrm{nm}\approx 103\mathrm{nm} \\ E=0.65\mathrm{eV} \\ \Rightarrow 0.65=\frac{hc}{\lambda } \\ \mathrm{or}\lambda =\frac{1242}{0.65}=1910.76\mathrm{nm}1911\mathrm{nm} \end{gathered}$$
Thus, the wavelengths of the radiations detected are 103 nm, 121 nm and 1911 nm.