Q153 / 160 · TS EAMCET 2009 · Physics
The work function of a certain metal is $3.31 \times 10^{-19} \mathrm{~J}$. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 Å$ is
$\begin{aligned}
& \text { (Given, } h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}, c=3 \times 10^8 \mathrm{~ms}^{-1} \text {, } \\
& e=1.6 \times 10^{-19} \mathrm{C} \text { ) }
\end{aligned}$
- A$2.48 \mathrm{eV}$
- Correct answerB$0.41 \mathrm{eV}$
- C$2.07 \mathrm{eV}$
- D$0.82 \mathrm{eV}$
Solution
Work function $W_0=3.31 \times 10^{-19} \mathrm{~J}$
Wavelength of incident radiation
$\begin{aligned}
& \lambda=5000 \times 10^{-10} \mathrm{~m} \\
& E=W_0+\mathrm{KE}
\end{aligned}$
(According to Einstein equation)
$\begin{aligned}
\frac{h c}{\lambda} & =3.31 \times 10^{-19}+\mathrm{KE} \\
\mathrm{KE} & =-3.31 \times 10^{-19}+\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} \\
& =-3.31 \times 10^{-19}+\frac{6.62 \times 3}{5} \times 10^{-19} \\
& =(-3.31 \times 1.324 \times 3) \times 10^{-19} \\
& =(3.972-3.31) \times 10^{-19} \\
& =0.662 \times 10^{-19} \mathrm{~J}
\end{aligned}$
$\begin{aligned} \Rightarrow E & =\frac{0.662 \times 10^{-19}}{1.6 \times 10^{-19}} \\ & =0.41 \mathrm{eV}\end{aligned}$
Hence, the correct answer is option: 'B'.