Dual Nature of Radiation and Matter

Step: Find the kinetic energy of the emitted electrons.
According to Einstein’s photoelectric equation, the kinetic energy of the emitted electron is given by;
\(K.E=h\nu-\phi\)
where, \(h\nu\) is the energy of the incident photons and \(\phi\) is the energy required for the emission of electrons.
\(\Rightarrow K.E=20~\text{eV}-13.6~\text{eV}~~~~~~~~~[\phi_\mathrm{H} = 13.6~\text{eV}]\)
\(\Rightarrow K.E = 6.4~\text{eV}\)
Hence, option (1) is the correct answer.
 

Hint: \(K.E_{\max }=h_f-\phi \)

Step: Find the velocity of the electron ejected from a silver surface by the ultraviolet light.
Given,
\( \lambda_0=3250 \times 10^{-10} \mathrm{~m}, ~~\lambda=2536 \times 10^{-10} \mathrm{~m} \)
\(\phi=\frac{{hc}}{\lambda_0}=\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{3250 \times 10^{-10}}=3.82 ~\mathrm{eV} \)
\({h} \nu=\frac{{hc}}{\lambda}=\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{2536 \times 10^{-10}}=4.89~ \mathrm{eV}\)
According to Einstein’s photoelectric equation,
\(K.E_{\max }=h_f-\phi \)
\(K.E_{\max }=(4.89-3.82) \mathrm{eV}=1.077 \mathrm{eV}\)
\(\frac{1}{2} m v^2=1.077 \times 1.6 \times 10^{-19} \)
\(v=\sqrt{\frac{2 \times 1.077 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \)
\(\Rightarrow v=0.6 \times 10^6{\mathrm{~m}}/{\mathrm{~s}}\)
Hence, option (1) is the correct answer.