Q155 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics
Energy required to remove an electron from aluminum surface is 4.2 eV . If light of wavelength $2000 Å$ falls on the surface, the velocity of the fastest electron ejected from the surface will be
- Correct answerA$8.4 \times 10^5 \mathrm{~ms}^{-1}$
- B$7.4 \times 10^5 \mathrm{~ms}^{-1}$
- C$6.4 \times 10^5 \mathrm{~ms}^{-1}$
- D$8.4 \times 10^6 \mathrm{~ms}^{-1}$
Solution
$\phi_{\mathrm{o}}=4.2 \mathrm{eV}, \lambda=2000 Å, \mathrm{v}_{\max }=?$
$\therefore \mathrm{E}=\frac{12400}{\lambda(\mathrm{in} Å)} \mathrm{eV}=\frac{12400}{2000}=6.2 \mathrm{eV}$
By Einstein's photoelectric equation,
$\mathrm{k}_{\max }=\mathrm{E}-\phi_0=6.2-4.2=2 \mathrm{eV}$
$\Rightarrow \frac{1}{2} \operatorname{mv}_{\max }^2=2 \times 1.6 \times 10^{-19}$
$\therefore \quad \mathrm{v}_{\max }=8.4 \times 10^5 \mathrm{~m} / \mathrm{s}$
Hence, the correct answer is option: 'A'.