A radiation of $3.8 \mathrm{eV}$ falls on a metal surface to produce photo electrons. These electrons are…
A radiation of $3.8 \mathrm{eV}$ falls on a metal surface to produce photo electrons. These electrons are made to enter a magnetic field of $2 \times 10^{-4} \mathrm{~T}$. If the radius of the largest circular path followed by these electrons is $30 \mathrm{~mm}$, then work function of the metal is
(Mass of electron $\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}$ )
$0.9 \mathrm{eV}$
$1.0 \mathrm{eV}$
$0.6 \mathrm{eV}$
$1.2 \mathrm{eV}$
Solution
$\begin{aligned} & K . E=\frac{q^2 B^2 r^2}{2 \mathrm{~m}} \\ \Rightarrow & K . E=\frac{\left(1.6 \times 10^{-19} \times 2 \times 10^{-4} \times 30 \times 10^{-3}\right)^2}{2 \times 9 \times 10^{-31}} \\ = & 5.06 \times 10^{-19} \mathrm{~J}=3.16 \mathrm{LeV} \\ & \text { So, work function }=3.8-3.16 \\ & \simeq 0.6 \mathrm{eV}\end{aligned}$