Q155 / 160 · TS EAMCET 2018 (04 May Shift 2) · Physics
An isolated lead ball is charged upon continuous irradiation by EM radiation of wavelength, $\lambda=221 \mathrm{~nm}$. The maximum potential attained by the lead ball, if its work function is $4.14 \mathrm{eV}$ is (take, $h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}$ )
- Correct answerA$1.49 \mathrm{~V}$
- B$2.67 \mathrm{~V}$
- C$3.14 \mathrm{~V}$
- D$0.51 \mathrm{~V}$
Solution
Given, $\lambda=2.21 \times 10^{-7} \mathrm{~m}$
So, energy of imparted by radiation in the lead ball,
$$
\begin{aligned}
E & =\frac{h c}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.21 \times 10^{-7}}=9 \times 10^{-19} \mathrm{~J} \\
\text { or } E & =\frac{9 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=5.63 \mathrm{eV}
\end{aligned}
$$
Work function of lead ball $=4 \cdot 14 \mathrm{eV}$
So, maximum energy attained by ball is
$$
5 \cdot 63-4 \cdot 14=1 \cdot 49 \mathrm{eV}
$$
Equivalent maximum potential is $1.49 \mathrm{~V}$.
Hence, the correct answer is option: 'A'.