Question

A beam of monochromatic light of wavelength λ ejects photoelectrons from a cesium surface (Φ = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of λ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.

Answer 1

Given:

Work function of cesium surface, ϕ = 1.9 eV

(a) Energy required to ionise a hydrogen atom in its ground state, E = 13.6 eV
From the Einstein's photoelectric equation,
$\frac{\mathrm{hc}}{\lambda }=E+\varphi$
Here,
h = Planck's constant
c = Speed of light
$\lambda$ = Wavelength of light
$$\begin{gathered} \therefore \frac{\mathrm{hc}}{\mathrm{\lambda }}-1.9=13.6 \\ \Rightarrow \frac{1240}{\mathrm{\lambda }}=15.5 \\ \Rightarrow \mathrm{\lambda }=\frac{1240}{15.5} \\ \Rightarrow \mathrm{\lambda }=80\mathrm{nm} \end{gathered}$$

(b) When the electron is excited from the states n₁ = 1 to n₂ = 2, energy absorbed $\left(E_1\right)$ is given by
$$\begin{gathered} E_1=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ {E}_1=13.6\left(1-\frac{1}{4}\right) \\ {E}_1=\frac{13.66\times 3}{4} \end{gathered}$$

For Einstein's photoelectric equation,

$$\begin{gathered} \therefore \frac{\mathrm{hc}}{\lambda }-1.9=\frac{13.6\times 3}{4} \\ \Rightarrow \frac{\mathrm{hc}}{\lambda }=\frac{13.6\times 3}{4}+1.9 \\ \Rightarrow \frac{1240}{\lambda }=10.2+1.9=12.1 \\ \Rightarrow \lambda =\frac{1240}{12.1} \\ \Rightarrow \lambda =102.47=102\mathrm{nm} \end{gathered}$$

(c) Excited atom will emit visible light if an electron jumps from the second orbit_​ to third orbit, i.e. from n₁ = 2​ to n₂ = 3. This is because Balmer series lies in the visible region.
Energy (E₂) of this transition is given by
$$\begin{gathered} E_2=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ {E}_2=13.6\left(\frac{1}{4}-\frac{1}{9}\right) \\ {E}_2=\frac{13.66\times 5}{36} \end{gathered}$$

For Einstein's photoelectric equation,
$$\begin{gathered} \frac{\mathrm{hc}}{\lambda }-1.9=\frac{13.6\times 5}{36} \\ \Rightarrow \frac{\mathrm{hc}}{\lambda }=\frac{13.6\times 5}{36}+1.9 \\ \Rightarrow \frac{1240}{\lambda }=1.88+1.9=3.78 \\ \Rightarrow \lambda =\frac{1240}{3.78} \\ \Rightarrow \lambda =328.04\mathrm{nm} \end{gathered}$$