Question
The work function of caesium is $2.14 eV$. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 V$.
✓
Answer
(a) For the cut-off or threshold frequency, the energy $h v_0$ of the incident radiation must be equal to work function $\phi_0$, so that
$
\begin{aligned}
v_0 & =\frac{\phi_0}{h}=\frac{2.14 eV }{6.63 \times 10^{-34} J s } \\
& =\frac{2.14 \times 1.6 \times 10^{-19} J }{6.63 \times 10^{-34} J s }=5.16 \times 10^{14} Hz
\end{aligned}
$
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy $e V_0$ by the retarding potential $V_0$. Einstein's Photoelectric equation is
$
\begin{aligned}
e V_0 & =h v-\phi_0=\frac{h c}{\lambda}-\phi_0 \\
\text { or, } \quad \lambda & =h c /\left(e V_0+\phi_0\right) \\
& =\frac{\left(6.63 \times 10^{-34} J s \right) \times\left(3 \times 10^8 m / s \right)}{(0.60 eV +2.14 eV )} \\
& =\frac{19.89 \times 10^{-26} J m }{(2.74 eV )} \\
\lambda & =\frac{19.89 \times 10^{-26} J m }{2.74 \times 1.6 \times 10^{-19} J }=454 nm
\end{aligned}
$
$
\begin{aligned}
v_0 & =\frac{\phi_0}{h}=\frac{2.14 eV }{6.63 \times 10^{-34} J s } \\
& =\frac{2.14 \times 1.6 \times 10^{-19} J }{6.63 \times 10^{-34} J s }=5.16 \times 10^{14} Hz
\end{aligned}
$
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy $e V_0$ by the retarding potential $V_0$. Einstein's Photoelectric equation is
$
\begin{aligned}
e V_0 & =h v-\phi_0=\frac{h c}{\lambda}-\phi_0 \\
\text { or, } \quad \lambda & =h c /\left(e V_0+\phi_0\right) \\
& =\frac{\left(6.63 \times 10^{-34} J s \right) \times\left(3 \times 10^8 m / s \right)}{(0.60 eV +2.14 eV )} \\
& =\frac{19.89 \times 10^{-26} J m }{(2.74 eV )} \\
\lambda & =\frac{19.89 \times 10^{-26} J m }{2.74 \times 1.6 \times 10^{-19} J }=454 nm
\end{aligned}
$
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