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The work function of metals is in the range of $2 \mathrm{eV}$ to $5 \mathrm{eV}$. Find which of the following wavelength of light cannot be used for photoelectric effect? (Consider, Plank constant $=4 \times 10^{-15} \mathrm{eVs},$ velocity of light $\left.=3 \times 10^{8} \mathrm{m} / \mathrm{s}\right)$
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Verified Answer
The correct answer is:
$650 \mathrm{nm}$
Given that, $2 \mathrm{eV} \leq \phi \leq 5 \mathrm{eV}$
Wavelength corresponding to minimum and maximum values of work function are
$$
\begin{aligned}
\lambda_{\max } &=\frac{h C}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^{8}}{2} \\
&=6 \times 10^{-7} \mathrm{m}=600 \mathrm{nm} \\
\text { and } \lambda_{\min } &=\frac{h C}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^{8}}{5} \\
&=2.4 \times 10^{-7} \mathrm{m}=240 \mathrm{nm}
\end{aligned}
$$
Clearly, wavelength range would be
$$
240 \leq \lambda \leq 600
$$
Thus, wavelength $650 \mathrm{nm}$ is not suitable for photoelectric effect.
Wavelength corresponding to minimum and maximum values of work function are
$$
\begin{aligned}
\lambda_{\max } &=\frac{h C}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^{8}}{2} \\
&=6 \times 10^{-7} \mathrm{m}=600 \mathrm{nm} \\
\text { and } \lambda_{\min } &=\frac{h C}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^{8}}{5} \\
&=2.4 \times 10^{-7} \mathrm{m}=240 \mathrm{nm}
\end{aligned}
$$
Clearly, wavelength range would be
$$
240 \leq \lambda \leq 600
$$
Thus, wavelength $650 \mathrm{nm}$ is not suitable for photoelectric effect.
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