Q90 / 120 · NEET 2012 (Mains) · Physics
Two radiations of photon energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$, successively illuminate a photosensitive metallic surface of work function $0.5 \mathrm{eV}$. The ratio of the maximum speeds of the emitted electrons is
- A$1: 4$
- Correct answerB$1: 2$
- C$1: 1$
- D$1: 5$
Solution
We have
$$
\frac{1}{2} m v_{\max }^2=E-\phi
$$
Here case (i)
$$
\begin{gathered}
\frac{1}{2} m v_{1 \max }^2=(1-0.5) \mathrm{eV} \\
\frac{1}{2} m v_{2 \max }^2=(2.5-0.5) \mathrm{eV}
\end{gathered}
$$
Hence, $\quad \frac{v_{1 \max }^2}{v_{2 \max }^2}=\frac{1}{4}$
$$
\frac{v_{1 \max }}{v_{2 \max }}=\frac{1}{2}
$$
Hence, the correct answer is option: 'B'.