KCET · Physics · Dual Nature of Matter
The photoelectric threshold wavelength for silver is \(\lambda_{0}\). The energy of the electron ejected from the surface of silver by an incident wavelength \(\lambda\left(\lambda < \lambda_{0}\right)\) will be
- A \(h c\left(\lambda_{0}-\lambda\right)\)
- B \(\frac{h c}{\lambda_{0}-\lambda}\)
- C \(\frac{h}{c}\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right)\)
- D \(h c\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right)\)
Answer & Solution
Correct Answer
(D) \(h c\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right)\)
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
E &=W+\mathrm{KE} \\
\mathrm{KE} &=E-W \\
&=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}} \\
&=h c\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]=h c\left[\frac{\lambda_{0}-\lambda}{\lambda_{0}}\right]
\end{aligned}
\]
\begin{aligned}
E &=W+\mathrm{KE} \\
\mathrm{KE} &=E-W \\
&=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}} \\
&=h c\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]=h c\left[\frac{\lambda_{0}-\lambda}{\lambda_{0}}\right]
\end{aligned}
\]
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