MHT CET · Physics · Dual Nature of Matter
For a photosensitive material, work function is ' \(\mathrm{W}_0\) ' and stopping potential is ' V '. The wavelength of incident radiation is (\(\mathrm{h}=\) Planck's constant, \(c=\) velocity of light, \(e=\) electronic charge)
- A \(\frac{\mathrm{h}^2 \mathrm{c}^2}{\mathrm{~W}_0+\mathrm{eV}}\)
- B \(\frac{\mathrm{hc}}{\mathrm{W}_0}\)
- C \(\frac{\mathrm{hcV}}{\mathrm{W}_0}\)
- D \(\frac{\mathrm{hc}}{\mathrm{W}_0+\mathrm{eV}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{hc}}{\mathrm{W}_0+\mathrm{eV}}\)
Step-by-step Solution
Detailed explanation
From Einstein's photoelectric equation, we can write,
\(\begin{array}{ll}
& \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_0 \\
\therefore \quad & \frac{\mathrm{hc}}{\lambda}=\mathrm{W}_0+\mathrm{eV} \\
\therefore \quad & \lambda=\frac{\mathrm{hc}}{\mathrm{~W}_0+\mathrm{eV}}
\end{array}\)
\(\begin{array}{ll}
& \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_0 \\
\therefore \quad & \frac{\mathrm{hc}}{\lambda}=\mathrm{W}_0+\mathrm{eV} \\
\therefore \quad & \lambda=\frac{\mathrm{hc}}{\mathrm{~W}_0+\mathrm{eV}}
\end{array}\)
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