Search any question & find its solution
Question:
Answered & Verified by Expert
Light of wavelength $180 \mathrm{nm}$ ejects photoelectron from a plate of a metal whose work function is $2 \mathrm{eV}$. If a uniform magnetic field of $5 \times 10^{-5} \mathrm{~T}$ is applied parallel to plate, what would be theradius of the path followed by electrons ejected normally from the plate with maximum energy?
Options:
Solution:
1669 Upvotes
Verified Answer
The correct answer is:
$0.149 \mathrm{~m}$
If $\mathrm{v}_{\max }$ is the speed of the fastest electron emitted from the metal surface, then
$$
\begin{aligned}
& \frac{h c}{\lambda}=W_{0}+\frac{1}{2} m v_{\max }^{2} \frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(180 \times 10^{-9}\right)} \\
&=2 \times\left(1.6 \times 10^{-19}\right)+\frac{1}{2}\left(9.1 \times 10^{-31}\right) v_{\max }^{2} \\
\therefore \quad & v=1.31 \times 10^{6} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
The radius of the electron is given by
$$
r=\frac{m v}{q B}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.31 \times 10^{6}\right)}{\left(1.6 \times 10^{-19}\right) \times\left(5 \times 10^{-9}\right)}=0.149 \mathrm{~m}
$$
$$
\begin{aligned}
& \frac{h c}{\lambda}=W_{0}+\frac{1}{2} m v_{\max }^{2} \frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(180 \times 10^{-9}\right)} \\
&=2 \times\left(1.6 \times 10^{-19}\right)+\frac{1}{2}\left(9.1 \times 10^{-31}\right) v_{\max }^{2} \\
\therefore \quad & v=1.31 \times 10^{6} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
The radius of the electron is given by
$$
r=\frac{m v}{q B}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.31 \times 10^{6}\right)}{\left(1.6 \times 10^{-19}\right) \times\left(5 \times 10^{-9}\right)}=0.149 \mathrm{~m}
$$
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.