If the work function for gold is \(8.155 \times 10^{-20} J\), what is the minimum frequency of light irradiated on the surface of gold that will emit photoelectrons?
Solution
Minimum frequency = threshold frequency

Work function,
\[
\begin{aligned}
W & =h \int_0 \\
\int_0 & =\frac{W}{h} \\
& =\frac{8.155 \times 10^{-20}}{6.63 \times 10^{-34}} \\
& =1.23 \times 10^{14} Hz
\end{aligned}
\]

Example

The graph below shows the variation of maximum kinetic energy of a photoelectron emitted from a metal surface with the frequency of light incident on the metal surface.

Determine the work function and threshold frequency of the metal.

Solution
\[
\begin{aligned}
E= & W+K_{\max } \\
W= & E-K_{\max } \\
= & h f-K_{\max } \\
= & \left(6.63 \times 10^{-34}\right)\left(12 \times 10^{14}\right) \\
& -\left(5.0 \times 10^{-20}\right) \\
= & 7.456 \times 10^{-19} J
\end{aligned}
\]

From,
\[
\begin{aligned}
W & =h \int_0 \\
f_0 & =\frac{W}{h} \\
& =\frac{7.456 \times 10^{-19}}{6.63 \times 10^{-34}} \\
& =1.125 \times 10^{15} Hz
\end{aligned}
\]

Example

A metal surface is illuminated by photons of energy \(4.0 eV\) and emits photoelectrons of maximum velocity \(2.0 \times 10^5 m s ^{-1}\). What is the maximum wavelength of the incident light wave that will emit electrons from the metal surface?
Solution
Energy of photon,
\[
\begin{aligned}
E & =4.0 eV \\
& =4.0\left(1.6 \times 10^{-19}\right) \\
& =6.4 \times 10^{-19} J
\end{aligned}
\]Maximum kinetic energy,
\[
\begin{aligned}
K_{\max } & =\frac{1}{2} m v_{\max }^2 \\
& =\frac{1}{2}\left(9.11 \times 10^{-31}\right)\left(2.0 \times 10^5\right)^2 \\
& =1.82 \times 10^{-20} J \\
E & =W+K_{\max } \\
W & =E-K_{\max } \\
& =\left(6.4 \times 10^{-19}\right)-\left(1.82 \times 10^{-20}\right) \\
& =6.218 \times 10^{-19} J
\end{aligned}
\]

From,
\[
\begin{array}{l}
\text { From, } W=h f_0 \\
=\frac{h c}{\lambda_{\max }} \\
\begin{aligned}
\lambda_{\max } & =\frac{h c}{W} \\
= & \frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}{ W } \\
= & \frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}{6.218 \times 10^{-19}} \\
= & 3.199 \times 10^{-7} m
\end{aligned}
\end{array}
\]

Example

Photoelectrons of maximum kinetic energy \(1.02 eV\) are emitted when a monochromatic light of wavelength \(550 nm\) is incident on a metal surface. The maximum kinetic energy becomes \(1.35 eV\) when a light of \(480 nm\) is used.
(a) Deduce the value of Planck’s constant.
(b) Determine the work function of the metal in units of \(eV\).
Solution
(a)
\[
\begin{array}{l}
E=W+K_{\max } \\
\frac{h c}{\lambda}=W+K_{\max } \\
\frac{h\left(3 \times 10^8\right)}{550 \times 10^{-9}}=W+1.02 eV \\
\frac{h\left(3 \times 10^8\right)}{480 \times 10^{-9}}=W+1.35 eV
\end{array}
\]
(2) – (1):
\[
\begin{array}{l}
\frac{h\left(3 \times 10^8\right)}{10^{-9}}\left(\frac{1}{480}-\frac{1}{550}\right) \\
=(1.35-1.02) eV \\
h=\frac{0.33 eV }{7.955 \times 10^{13}}
\end{array}
\]Converting energy in \(eV\) to \(J\) :
\[
\begin{aligned}
h & =\frac{(0.33)\left(1.6 \times 10^{-19}\right)}{7.955 \times 10^{13}} \\
& =6.64 \times 10^{-34} J s ^{-1} \ldots
\end{aligned}
\]
(b) (3) into (1):
\[
\begin{array}{l}
\frac{\left(6.64 \times 10^{-34}\right)\left(3 \times 10^8\right)}{550 \times 10^{-9}} \\
= W +1.02 eV
\end{array}
\]

Converting energy in \(J\) to \(eV\),
\[
\begin{array}{l}
\frac{\left(6.64 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\left(550 \times 10^{-9}\right)\left(1.6 \times 10^{-19}\right)} \\
=W+1.02 eV \\
2.26 eV = W +1.02 eV
\end{array}
\]

Therefore, work function, \(W=1.24 eV\)

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