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 emitted photoelectrons are \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2}\) respectively. The work function of the photosensitive surface is
- A \(\frac{\lambda_{2} E_{2}-\lambda_{1} E_{1}}{\lambda_{2}-\lambda_{1}}\)
- B \(\frac{\lambda_{1} E_{1}+\lambda_{2} E_{2}}{\lambda_{2}+\lambda_{1}}\)
- C \(\frac{\lambda_{1} E_{1}-\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}\)
- D \(\frac{\lambda_{2} E_{1}+\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda_{1} E_{1}-\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}\)
Step-by-step Solution
Detailed explanation
Using photoclutic effect
\(\frac{h c_{1}}{\lambda_{1}}=\phi+k_{1}-(D)\) ...(i)
and \(\frac{h_{1}}{r_{2}}=\phi k k_{2}\) ...(ii)
form (i) and (i)
\(\frac{\frac{h n_{1}}{\lambda_{1}}}{\frac{h_{2}}{\lambda_{2}}}=\frac{\min (\eta) \text { an }}{\phi+k_{1}}\)
\(\Rightarrow \frac{1}{d_{1}} \times \frac{d_{2}}{1}=\frac{\phi+k_{1}}{\phi+k_{2}}\)
\(\Rightarrow \frac{d_{2}}{n_{1}}=\frac{\phi+4}{\phi+b_{2}}\)
\(\Rightarrow \lambda_{2} \phi+k_{2} d_{2}=\lambda_{1} \phi+d_{1} k_{1}\)
\(\Rightarrow d_{2} \phi-\lambda_{1} \phi=\lambda_{1} k_{1}-k_{2} d_{2}\)
\(\therefore=\frac{d_{1} k_{1}-k_{2} d_{2}}{d_{2}-d_{1}}\)
\(\frac{h c_{1}}{\lambda_{1}}=\phi+k_{1}-(D)\) ...(i)
and \(\frac{h_{1}}{r_{2}}=\phi k k_{2}\) ...(ii)
form (i) and (i)
\(\frac{\frac{h n_{1}}{\lambda_{1}}}{\frac{h_{2}}{\lambda_{2}}}=\frac{\min (\eta) \text { an }}{\phi+k_{1}}\)
\(\Rightarrow \frac{1}{d_{1}} \times \frac{d_{2}}{1}=\frac{\phi+k_{1}}{\phi+k_{2}}\)
\(\Rightarrow \frac{d_{2}}{n_{1}}=\frac{\phi+4}{\phi+b_{2}}\)
\(\Rightarrow \lambda_{2} \phi+k_{2} d_{2}=\lambda_{1} \phi+d_{1} k_{1}\)
\(\Rightarrow d_{2} \phi-\lambda_{1} \phi=\lambda_{1} k_{1}-k_{2} d_{2}\)
\(\therefore=\frac{d_{1} k_{1}-k_{2} d_{2}}{d_{2}-d_{1}}\)
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