Question

Threshold wavelength for a metal surface is 2000 \AA. On incidence of radiation of 1000 \AA, the kinetic energy of emitted photoelectrons will be

• A. 6.2 eV
• B. 12.4 eV
• C. 3.6 eV
• D. 2.6 eV

Hint:

Use \(E(\text{eV}) \approx \dfrac{1240}{\lambda(\text{nm})}\). Then \(K_{\max}=E-\Phi\) with \(\Phi = \dfrac{1240}{\lambda_0}\).

Answer 1

The Correct Option is A

Work function \(\Phi=\dfrac{hc}{\lambda_0}\) with \(\lambda_0=2000~\text{\AA}=200~\text{nm}\). Using \(hc=1240~\text{eV}\cdot\text{nm}\): \(\Phi=\dfrac{1240}{200}=6.2~\text{eV}\).
Photon energy at \(\lambda=1000~\text{\AA}=100~\text{nm}\): \(E_\gamma=\dfrac{1240}{100}=12.4~\text{eV}\).
Maximum kinetic energy: \(K_{\max}=E_\gamma-\Phi=12.4-6.2=6.2~\text{eV}\).