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Light of two different frequencies whose photons have energies $1.3 \mathrm{eV}$ and $2.8 \mathrm{eV}$ respectively, successfully illuminate a metallic surface whose work function is $0.8 \mathrm{eV}$. The ratio of maximum speeds of emitted electrons will be
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The correct answer is:
$1: 2$
For maximum speed of the photo electrons, $\frac{1}{2} m v^2=E_{\mathrm{P}}-\phi$ where $\phi=0.8 \mathrm{eV}$ is the work function of the metal. Energy of photon $E_1=1.3 \mathrm{eV}$
$$
\therefore \frac{1}{2} m v_1^2=1.3-0.8 \mathrm{eV}
$$
Ratio of maximum speeds of the electrons can obtained by taking ratio of equation (1) and (2),
$\therefore \frac{v_1^2}{v_2^2}=\frac{0.5}{2}=\frac{1}{4}$
Or, $\frac{v_1}{v_2}=\frac{1}{2}$
$$
\therefore \frac{1}{2} m v_1^2=1.3-0.8 \mathrm{eV}
$$
Ratio of maximum speeds of the electrons can obtained by taking ratio of equation (1) and (2),
$\therefore \frac{v_1^2}{v_2^2}=\frac{0.5}{2}=\frac{1}{4}$
Or, $\frac{v_1}{v_2}=\frac{1}{2}$
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