Light of wavelength \(488 \mathrm{~nm}\) produced by an Agron laser is used in the photoelectric effect.…
Light of wavelength \(488 \mathrm{~nm}\) produced by an Agron laser is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping potential of the photoelectrons is \(0.38 \mathrm{~V}\). The work function of the cathode material is
\(2.16 \mathrm{eV}\)
\(216 \mathrm{eV}\)
\(21.6 \mathrm{eV}\)
\(0.216 \mathrm{eV}\)
Solution
Given, light of wavelength, \(\lambda=488 \mathrm{~nm}\) and stopping potential, of photoelectrons, \(V=0.38 \mathrm{~V}\) As, energy of photon
\(E=\frac{h c}{\lambda}=\frac{1250}{488} \mathrm{eV}\)
So, \(E=2.56 \mathrm{eV}\)
\(\because\) Maximum kinetic energy of a photo-electron
\(\mathrm{KE}_{\max }=e V_0\)
Where, \(V_0=\) stopping potential
\(\therefore \quad \mathrm{KE}_{\max }=e V_0=0.38 \mathrm{eV}\)
From Einstein's photoelectric equation,
\(\begin{aligned}
& E=\mathrm{KE}_{\max }+W_0 \\
& \Rightarrow \quad W_0=E-\mathrm{KE}_{\max } \\
& \quad=2.56-0.38=2.18 \approx 2.16 \mathrm{eV}
\end{aligned}\)
Hence, the work function of cathode material is \(2.16 \mathrm{eV}\).
So, the correct option is (a).