All electrons ejected from a metallic surface by incident light of wavelength \(400 \mathrm{~nm}\) travelled…
All electrons ejected from a metallic surface by incident light of wavelength \(400 \mathrm{~nm}\) travelled \(1 \mathrm{~m}\) in the direction of uniform electric field of \(2 \mathrm{NC}^{-1}\) and came to rest. The work function of the surface is
\(1.1 \mathrm{eV}\)
\(2.2 \mathrm{eV}\)
\(3.1 \mathrm{eV}\)
\(5.1 \mathrm{eV}\)
Solution
Given, wavelength of incident radiation,
\(\lambda=400 \mathrm{~nm}=4 \times 10^{-7} \mathrm{~m}\)
electric field, \(E=2 \mathrm{~N} / \mathrm{C}\) and distance, \(s=1 \mathrm{~m}\)
\(\therefore\) Energy of the incident light,
\(\begin{aligned}
E & =\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \\
& =4.95 \times 10^{-19} \mathrm{~J}=\frac{4.95 \times 10^{19}}{1.6 \times 10^{-19}} \\
\mathrm{~J} & =3.09 \simeq 3 \mathrm{leV}
\end{aligned}\)
If \(a\) be the retardation of emitted electrons in the electric field,
then, \(\quad a=\frac{q E}{m}\)
initial speed \(u\) of emitted electron is calculated as
\(v^2=u^2-2 a s \Rightarrow 0=u^2-2 a \times 1 \Rightarrow u^2=2 a=\frac{2 q E}{m}\)
\(\therefore\) Maximum kinetic energy of the electron,
\(\begin{aligned}
& K_{\max }=\frac{1}{2} m u^2=m \cdot \frac{2 q E}{m}=q E=1.6 \times 10^{-19} \times 2 \\
& K_{\max }=3.2 \times 10^{-19} \mathrm{~J}=2 \mathrm{eV}
\end{aligned}\)
\(\therefore\) Work function of the surface,
\(W_0=E-K_{\max }=3.1-2=1.1 \mathrm{eV}\)