Light of two different frequencies whose photons have energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$…

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
  1. $1: 2$
  2. $1: 1$
  3. $1: 5$
  4. $1: 4$

Solution

Kinetic energy $\mathrm{KE}=\phi-\phi_0$ Here, $\mathrm{KE}_1=1-0.5=0.5 \mathrm{eV}$ $\begin{aligned} & \mathrm{KE}_2=2.5-0.5=2 \mathrm{eV} \\ & \therefore \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{0.5}{2}=\frac{1}{4} \\ & \text { or } \quad \frac{y_1^2}{y_2^2}=\frac{1}{4} \\ & \text { or } \quad \frac{y_1}{y_2}=\sqrt{\frac{1}{4}}=\frac{1}{2} \end{aligned}$

Asked in: NEET 2011 (Screening)

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