Question 43: Dual Nature of Matter and Radiation

Question

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \mathrm{~V}$. This potential drops to $0.6 \mathrm{~V}$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $\frac{h c}{e}=1.24 	imes$ $10^{-6} \mathrm{JmC}^{-1}$.]

Accepted Answer

By Einstein's photoelectric equation, we have:
$

\mathrm{KE}  =h v-\phi$
$
h v-\phi  =6 \mathrm{eV} $
$
\frac{h c}{\lambda}-\phi  =6 \mathrm{eV} $ .........(i)
$
\frac{h c}{4 \lambda}-\phi  =0.6 \mathrm{eV} $
$
\frac{3 h c}{4 \lambda}  =5.4 \mathrm{eV}$
$$ \Rightarrow $$ $\lambda=\frac{3 h c}{5.4 \mathrm{eV} 	imes 4} $
$$ \Rightarrow $$ $
\lambda =\frac{3 	imes 1.24 	imes 10^{-6}}{4 	imes 5.4}=1.72 	imes 	imes^{-7} \mathrm{~m}$ ........(ii)
From equation (i) and (ii), we conclude
$\frac{h c}{1.72 	imes 10^{-7}} 	imes \frac{1}{1.6 	imes 10^{-19}}-\phi=6 \mathrm{eV}$
on subsituting all the values we get
$$
\begin{aligned}
& \phi=(7.27-6) \\
&\Rightarrow \phi=1.27 \mathrm{eV} \\
&\Rightarrow \phi=1.20 \mathrm{eV}
\end{aligned}
$$

Answer

$1.72 	imes 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}$

Answer

$1.72 	imes 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}$

Answer

$3.78 	imes 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}$

Answer

$3.78 	imes 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}$

Dual Nature of Matter and Radiation

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