MHT CET · Physics · Dual Nature of Matter
Photons of wavelength \(\lambda\) ' are incident on the cathode of a photocell. Electrons are emitted from the cathode surface. The de-Broglie wavelength of the emitted electrons is (work function is negligible)
\((\mathrm{c}=\) velocity of light, \(\quad \mathrm{h}=\) Planck's constant, \(\quad \mathrm{m}=\) mass of electron \()\)
- A \(\sqrt{\frac{\mathrm{mc}}{2 \mathrm{~h} \lambda}}\)
- B \(\sqrt{\frac{\mathrm{h} \lambda}{2 \mathrm{mc}}}\)
- C \(\sqrt{\frac{2 \mathrm{~h} \lambda}{\mathrm{mc}}}\)
- D \(\sqrt{\frac{\mathrm{mh}}{\lambda \mathrm{c}}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{\mathrm{h} \lambda}{2 \mathrm{mc}}}\)
Step-by-step Solution
Detailed explanation
Work function is negligible
\(\begin{array}{l}
\therefore \frac{1}{2} \mathrm{mV}^{2}=\frac{\mathrm{hc}}{\lambda} \\
\therefore \frac{1}{2} \frac{\mathrm{m}^{2} \mathrm{~V}^{2}}{\mathrm{~m}}=\frac{\mathrm{hc}}{\lambda} \\
\therefore \mathrm{m}^{2} \mathrm{~V}^{2}=\frac{2 \mathrm{mhC}}{\lambda} \\
\therefore \mathrm{mV}=\sqrt{\frac{2 \mathrm{mhc}}{\lambda}}
\end{array}\)
\(\begin{aligned}
\text {de Broglie wavelength } &=\frac{\mathrm{h}}{\mathrm{mV}} \\
&=\mathrm{h} \sqrt{\frac{\lambda}{2 \mathrm{mhC}}}=\sqrt{\frac{\mathrm{h} \lambda}{2 \mathrm{mC}}}
\end{aligned}\)
\(\begin{array}{l}
\therefore \frac{1}{2} \mathrm{mV}^{2}=\frac{\mathrm{hc}}{\lambda} \\
\therefore \frac{1}{2} \frac{\mathrm{m}^{2} \mathrm{~V}^{2}}{\mathrm{~m}}=\frac{\mathrm{hc}}{\lambda} \\
\therefore \mathrm{m}^{2} \mathrm{~V}^{2}=\frac{2 \mathrm{mhC}}{\lambda} \\
\therefore \mathrm{mV}=\sqrt{\frac{2 \mathrm{mhc}}{\lambda}}
\end{array}\)
\(\begin{aligned}
\text {de Broglie wavelength } &=\frac{\mathrm{h}}{\mathrm{mV}} \\
&=\mathrm{h} \sqrt{\frac{\lambda}{2 \mathrm{mhC}}}=\sqrt{\frac{\mathrm{h} \lambda}{2 \mathrm{mC}}}
\end{aligned}\)
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