Q134 / 180 · KCET 2021 · Physics
The work-function of a metal is $1 \mathrm{eV}$. Light of wavelength $3000 Å$ is incident on this metal surface. The velocity of emitted photoelectrons will be
- A$10 \mathrm{~ms}^{-1}$
- B$1 \times 10^{3} \mathrm{~ms}^{-1}$
- C$1 \times 10^{4} \mathrm{~ms}^{-1}$
- Correct answerD$1 \times 10^{6} \mathrm{~ms}^{-1}$
Solution
Given, Work-function, $\phi_{0}=\mathrm{leV}=1.6 \times 10^{-19} \mathrm{~J}$ Wavelength, $\lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m}$
As, we know that,
$E=h v=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{3000 \times 10^{-10}}$
$=6.6 \times 10^{-19} \mathrm{~J}$
Also, $E=\phi_{0}+\mathrm{KE}$
$\Rightarrow \quad \mathrm{KE}=E-\phi_{0} \Rightarrow \frac{1}{2} m v^{2}=E-\phi_{0}$
$v^{2}=\frac{\left(6.6 \times 10^{-19}-1.6 \times 10^{-19}\right) \times 2}{9.1 \times 10^{-31}}$
$=\frac{5 \times 10^{-19} \times 2}{9.1 \times 10^{-31}}=1.1 \times 10^{12}$
$v=1.01 \times 10^{6}$
$\simeq 1 \times 10^{6} \mathrm{~m} / \mathrm{s}$
Hence, the correct answer is option: 'D'.