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A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$, then the radius $r$ is given by, (in the usual notation)
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Verified Answer
The correct answer is:
$\sqrt{2 m\left(E-W_0\right) e B}$
From Einstein's equation
$\begin{gathered}
E=W_0+\frac{1}{2} m v^2 \\
\sqrt{\frac{2\left(E-W_0\right)}{m}}=v
\end{gathered}$
or A charged particle placed in uniform magnetic field experience a force
$F=\frac{m v^2}{r}$
$\begin{aligned} \text { or } & e v B & =\frac{m v^2}{r} \\ \text { or } & r & =\frac{m v}{e B}\end{aligned}$
$\begin{array}{ll}\text { or } & r=\frac{m \sqrt{\frac{2\left(E-W_0\right)}{m}}}{e B} \\ \Rightarrow & r=\frac{\sqrt{2 m\left(E-W_0\right)}}{e B}\end{array}$
$\begin{gathered}
E=W_0+\frac{1}{2} m v^2 \\
\sqrt{\frac{2\left(E-W_0\right)}{m}}=v
\end{gathered}$
or A charged particle placed in uniform magnetic field experience a force
$F=\frac{m v^2}{r}$
$\begin{aligned} \text { or } & e v B & =\frac{m v^2}{r} \\ \text { or } & r & =\frac{m v}{e B}\end{aligned}$
$\begin{array}{ll}\text { or } & r=\frac{m \sqrt{\frac{2\left(E-W_0\right)}{m}}}{e B} \\ \Rightarrow & r=\frac{\sqrt{2 m\left(E-W_0\right)}}{e B}\end{array}$
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