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A light of wavelength $310 \mathrm{~nm}$ is used in a photoelectric experiment. The metal electrode of work function of $2.5 \mathrm{eV}$ is used in the experiment. The stopping potential for the photoelectrons will be (assume, $h c=1240 \mathrm{eV}-\mathrm{nm}$ )
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The correct answer is:
1.5 V
Work-function of metal surface, $\phi_0=25 \mathrm{eV}$
Wavelength, $\lambda=310 \mathrm{~nm} \Rightarrow h c=1240 \mathrm{eV}-\mathrm{nm}$
According to Einstein's photoelectric equation,
$\begin{aligned}
e V_0 & =\frac{h c}{\lambda}-\phi_0 \Rightarrow e V_0=\frac{1240 \mathrm{eV}-\mathrm{nm}}{310 \mathrm{~nm}}-2.5 \mathrm{eV} \\
\Rightarrow \quad V_0 & =\frac{1240}{310}-25=4-25=1.5 \mathrm{~V}
\end{aligned}$
Wavelength, $\lambda=310 \mathrm{~nm} \Rightarrow h c=1240 \mathrm{eV}-\mathrm{nm}$
According to Einstein's photoelectric equation,
$\begin{aligned}
e V_0 & =\frac{h c}{\lambda}-\phi_0 \Rightarrow e V_0=\frac{1240 \mathrm{eV}-\mathrm{nm}}{310 \mathrm{~nm}}-2.5 \mathrm{eV} \\
\Rightarrow \quad V_0 & =\frac{1240}{310}-25=4-25=1.5 \mathrm{~V}
\end{aligned}$
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