Question
The work function of caesium is 2.14 eV . Find
(a) The threshold frequency for caesium.
(b) The wavelength of incident light if photocurrent is brought to zero by a stopping potential of $0.60 V .\left( h =6.625 \times 10^{-34} Js \right)$

Answer


$
\begin{array}{c}
\varphi_0=2.14 eV \\
V_0=0.60 V \\
v_0=? \\
\lambda=?
\end{array}
$
→ (a)The cut off or threshold frequency
$
\begin{array}{l}
\varphi_0=h v_0 \\
\therefore v_0=\frac{\phi_0}{h} \\
=\frac{2.14 \times 1.6 \times 10^{-19}}{6.625 \times 10^{-34}} \\
v_0=0.5163 \times 10^{15} Hz \\
v_0=5.16 \times 10^{14} Hz
\end{array}
$
→ (b) Einstein's photoelectric equation is
$\begin{array}{l} K _{\max }=h v-\varphi_0 \\ \text { but } K _{\max }=e V_0 \\ \therefore e V_0=\frac{h c}{\lambda}-\varphi_0( C =v \lambda) \\ \therefore \frac{h c}{\lambda}=e V_0+\varphi_0 \\ \therefore \frac{h c}{\lambda}=\left(1.6 \times 10^{-19} \times 0.60\right)+\left(2.14 \times 1.6 \times 10^{-19}\right) \\ \therefore \frac{h c}{\lambda}=1.6 \times 10^{-19}(0.60+2.14)\end{array}$
$\therefore \frac{h c}{\lambda}=4.384 \times 10^{-19}$
$\therefore \lambda=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{4.384 \times 10^{-19}}$
$\begin{array}{l}\therefore \lambda=4.54 \times 10^{-7} m \\ =454 \times 10^{-9} m \\ \lambda=454 nm\end{array}$

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