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MHT CET · Physics · Dual Nature of Matter

When a photosensitive surface is irradiated by lights of wavelengths ' \(\lambda_1\) ' and ' \(\lambda_2\) ', kinetic energies of the emitted photoelectrons is ' \(E_1\) ' and ' \(E_2\) ' respectively. The work function of the photosensitive surface is

  1. A \(\frac{\left(E_2 \lambda_2-E_1 \lambda_1\right)}{\left(\lambda_2-\lambda_1\right)}\)
  2. B \(\frac{\left(E_1 \lambda_1+E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}\)
  3. C \(\frac{\left(E_1 \lambda_1-E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}\)
  4. D \(\frac{\left(E_2 \lambda_2+E_1 \lambda_1\right)}{\left(\lambda_1-\lambda_2\right)}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\left(E_1 \lambda_1-E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}\)

Step-by-step Solution

Detailed explanation

From Einstein's photoelectric equation,
\(\begin{aligned}
& \mathrm{E}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \\
\therefore \quad & \mathrm{E}_1 \lambda_1=\mathrm{hc}-\mathrm{W}_0 \lambda_1 \\
& \mathrm{E}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\
\therefore \quad & \mathrm{hc}=\mathrm{E}_1 \lambda_1+\mathrm{W}_0 \lambda_1...(i)
\end{aligned}\)
\(\begin{array}{ll}
\therefore \quad & \mathrm{E}_2 \lambda_2=\mathrm{hc}-\mathrm{W}_0 \lambda_2 \\
& \Rightarrow \mathrm{hc}=\mathrm{E}_2 \lambda_2+\mathrm{W}_0 \lambda_2...(ii)
\end{array}\)
From equations (i) and (ii),
\(\begin{array}{ll}
& \mathrm{E}_1 \lambda_1+\mathrm{W}_0 \lambda_1=\mathrm{E}_2 \lambda_2+\mathrm{W}_0 \lambda_2 \\
\therefore \quad & \mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2=\mathrm{W}_0\left(\lambda_2-\lambda_1\right) \\
\therefore \quad & \mathrm{W}_0=\frac{\mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2}{\lambda_2-\lambda_1}
\end{array}\)