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The work function of a surface of a photosensitive material is $6.2 \mathrm{eV}$. The wavelength of the incident radiation for which the stopping potential is $5 \mathrm{~V}$ lies in the
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The correct answer is:
Ultraviolet region
$$
\begin{aligned}
& \mathrm{eV}_0=E-\phi \\
& E=\mathrm{eV}_0+\phi \\
& =5 \mathrm{eV}+6.2 \mathrm{eV} \\
& =11.2 \mathrm{eV} \\
& \therefore \quad \lambda=\left(\frac{12400}{11.2}\right) Å=1000 Å \\
& \Rightarrow \text { hence lies in ultraviolet region. } \\
&
\end{aligned}
$$
\begin{aligned}
& \mathrm{eV}_0=E-\phi \\
& E=\mathrm{eV}_0+\phi \\
& =5 \mathrm{eV}+6.2 \mathrm{eV} \\
& =11.2 \mathrm{eV} \\
& \therefore \quad \lambda=\left(\frac{12400}{11.2}\right) Å=1000 Å \\
& \Rightarrow \text { hence lies in ultraviolet region. } \\
&
\end{aligned}
$$
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