Q144 / 200 · NEET 2011 (Screening) · Physics
Light of two different frequencies whose photons have energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$ respectively illuminate a metallic surface whose work function is $0.5 \mathrm{eV}$ successively. Ratio of maximum speeds of emitted electrons will be
- Correct answerA$1: 2$
- B$1: 1$
- C$1: 5$
- D$1: 4$
Solution
Kinetic energy
$$
\mathrm{KE}=\phi-\phi_0
$$
Here, $\mathrm{KE}_1=1-0.5=0.5 \mathrm{eV}$
$$
\begin{aligned}
& \mathrm{KE}_2=2.5-0.5=2 \mathrm{eV} \\
& \therefore \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{0.5}{2}=\frac{1}{4} \\
& \text { or } \quad \frac{y_1^2}{y_2^2}=\frac{1}{4} \\
& \text { or } \quad \frac{y_1}{y_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.