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Photons of energy $2.4 \mathrm{eV}$ and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is
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The correct answer is:
$2.1 \mathrm{eV}$
Let $\phi_0=$ work-function of metal.
Then, according to Einstein's photoelectric equation,
$K_{\max }=\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0$
When wavelength is reduced by $50 \%$ i.e, $\lambda^{\prime}=\frac{\lambda}{2}$, then maximum velocity of emitted electrons is $3 v$.
From Eqs. (i) and (ii), we have
$\begin{aligned}
& 9\left(\frac{h c}{\lambda}-\phi_0\right)=\frac{2 h c}{\lambda}-\phi_0 \\
& 7 \frac{h c}{\lambda}=8 \phi_0 \Rightarrow \phi_0=\frac{7}{8} \frac{h c}{\lambda}
\end{aligned}$
Here, photon energy is given as $\frac{h c}{\lambda}=2.4 \mathrm{eV}$
So, work-function is $\phi_0=\frac{7}{8} \times 2.4=2.1 \mathrm{eV}$
Then, according to Einstein's photoelectric equation,
$K_{\max }=\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0$
When wavelength is reduced by $50 \%$ i.e, $\lambda^{\prime}=\frac{\lambda}{2}$, then maximum velocity of emitted electrons is $3 v$.
From Eqs. (i) and (ii), we have
$\begin{aligned}
& 9\left(\frac{h c}{\lambda}-\phi_0\right)=\frac{2 h c}{\lambda}-\phi_0 \\
& 7 \frac{h c}{\lambda}=8 \phi_0 \Rightarrow \phi_0=\frac{7}{8} \frac{h c}{\lambda}
\end{aligned}$
Here, photon energy is given as $\frac{h c}{\lambda}=2.4 \mathrm{eV}$
So, work-function is $\phi_0=\frac{7}{8} \times 2.4=2.1 \mathrm{eV}$
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