Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1402 Upvotes
Verified Answer
The correct answer is:
$h c\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right)$
$$
\begin{aligned}
E &=W+\mathrm{KE} \\
\mathrm{KE} &=E-W \\
&=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}} \\
&=h c\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]=h c\left[\frac{\lambda_{0}-\lambda}{\lambda_{0}}\right]
\end{aligned}
$$
\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}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.