The work function for caesium atom is 1.9 eV . Calculate (a) the …

Question

The work function for caesium atom is 1.9 eV . Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm , calculate the kinetic energy and the velocity of the rejected photoelectron.

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Answer

Work function of Cs atom $W _0=1.9 eV$
$
\begin{array}{l}
=1.9 \times 1.6 \times 10^{-19} J \\
\quad\left[1 eV=1.6 \times 10^{-19} J\right] \\
=3.04 \times 10^{-19} J
\end{array}
$
(a) $\quad W _0=\frac{h c}{\lambda_0}$
$\lambda_0=\frac{h c}{W_0}=\frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{3.04 \times 10^{-19} J}$
$\begin{aligned} \lambda_0 & =6.538 \times 10^{-7} m=6.54 \times 10^{-7} m \\ & =654 nm\end{aligned}$
Hence threshold wavelength of emitted radiation
$
=654 nm
$
(b) Threshold frequency
$\begin{aligned} v_0 & =\frac{c}{\lambda_0}=\frac{3 \times 10^8 ms^{-1}}{6.538 \times 10^{-7} m} \\ & =4.588 \times 10^{14} s^{-1} \\ & =4.59 \times 10^{14} s^{-1}\end{aligned}$
(c) $500 nm\left(500 \times 10^{-9} m\right)$ energy of photon (K.E.)
$
\begin{array}{l}
E=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{500 \times 10^{-9} m} \\
E=3.97 \times 10^{-19}
\end{array}
$
K.E. of emitted photoelectron
$
=h v-h v_0
$
$\begin{aligned} & =(3.97-3.04) \times 10^{-19} J \\ \text { K.E. } & =0.93 \times 10^{-19} J \\ & =9.30 \times 10^{-20} J \\ \text { K.E. } & =\frac{1}{2} m v ^2\end{aligned}$
Velocity of photoelectron $(v)=\sqrt{\frac{2 \times K . E .}{ m }}$
$=\sqrt{\frac{2 \times 9.30 \times 10^{-20} J}{9.1 \times 10^{-31} kg}}$
$v =4.52 \times 10^5 ms^{-1}$

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