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In a photoelectric effect experiment the cathode metal is exposed to light of wavelength $600 \mathrm{~nm}$. The maximum kinetic energy of the ejected electron doubles when light of wavelength $400 \mathrm{~nm}$ is used. The work function of the cathode metal is approximately. [Use $h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}, c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ]
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The correct answer is:
$1.02 \mathrm{eV}$
Wavelength of light, $\lambda_1=600 \mathrm{~nm}$
If the work-function is $\phi$ and the kinetic energy of electron is $E_K=\frac{1}{2} m v^2$, then we can write,
Also, when the wavelength, $\lambda_2=400 \mathrm{~nm}$, Kinetic energy of electron is $2 E_K=m v^2$
Multiplying Eq. (i) by 2 , we get
By subtracting Eq (iii) from Eq. (ii), we get
$$
\begin{aligned}
& \frac{h c}{\lambda_2}-\frac{2 h c}{\lambda_1}=m v^2+\phi-m v^2-2 \phi \\
& \phi=h c\left(\frac{2}{\lambda_1}-\frac{1}{\lambda_2}\right) \\
&=6.626 \times 10^{-34} \times 3 \times 10^8\left(\frac{2}{600}-\frac{1}{400}\right) \times 10^9 \\
&=6.626 \times 10^{-34} \times 3 \times 10^8\left(\frac{1}{300}-\frac{1}{400}\right) \times 10^9 \\
& \phi=1.02 \mathrm{eV}
\end{aligned}
$$
If the work-function is $\phi$ and the kinetic energy of electron is $E_K=\frac{1}{2} m v^2$, then we can write,
Also, when the wavelength, $\lambda_2=400 \mathrm{~nm}$, Kinetic energy of electron is $2 E_K=m v^2$
Multiplying Eq. (i) by 2 , we get
By subtracting Eq (iii) from Eq. (ii), we get
$$
\begin{aligned}
& \frac{h c}{\lambda_2}-\frac{2 h c}{\lambda_1}=m v^2+\phi-m v^2-2 \phi \\
& \phi=h c\left(\frac{2}{\lambda_1}-\frac{1}{\lambda_2}\right) \\
&=6.626 \times 10^{-34} \times 3 \times 10^8\left(\frac{2}{600}-\frac{1}{400}\right) \times 10^9 \\
&=6.626 \times 10^{-34} \times 3 \times 10^8\left(\frac{1}{300}-\frac{1}{400}\right) \times 10^9 \\
& \phi=1.02 \mathrm{eV}
\end{aligned}
$$
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