Q167 / 180 · KCET 2015 · Physics
Light of two different frequencies whose photons have energies \( 1 \mathrm{eV} \) and \( 2.5 \mathrm{eV} \) respectively,
successively illuminate a metallic surface whose work function is \( 0.5 \mathrm{eV} \). Ratio of maximum
speeds of emitted electrons will be
successively illuminate a metallic surface whose work function is \( 0.5 \mathrm{eV} \). Ratio of maximum
speeds of emitted electrons will be
- A$1: 5$
- B\( 1: 4 \)
- Correct answerC\( 1: 2 \)
- D1:1
Solution
Kinetic energy, \( K h=f W \)
where \( W \) is work function; \( f \) is frequency of incident photon.
For photon of energy
\( 1 e V, K_{1}=1 e V-0.5 e V=0.5 e V \)
For photon of energy
\( 2.5 \mathrm{eV}, K_{2}=2.5 \mathrm{eV}-0.5 \mathrm{eV}=2.0 \mathrm{eV} \)
Taking ratio, \( \frac{K_{1}}{K_{2}}=\frac{0.5 \mathrm{eV}}{2.0 \mathrm{eV}} \)
\( \Rightarrow \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}=\frac{0.5}{2.0} \Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4} \Rightarrow \frac{v_{1}}{v_{2}}=\frac{1}{2} \)
Therefore, ratio of maximum speeds of emitted electrons will be \( 1: 2 \).
where \( W \) is work function; \( f \) is frequency of incident photon.
For photon of energy
\( 1 e V, K_{1}=1 e V-0.5 e V=0.5 e V \)
For photon of energy
\( 2.5 \mathrm{eV}, K_{2}=2.5 \mathrm{eV}-0.5 \mathrm{eV}=2.0 \mathrm{eV} \)
Taking ratio, \( \frac{K_{1}}{K_{2}}=\frac{0.5 \mathrm{eV}}{2.0 \mathrm{eV}} \)
\( \Rightarrow \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}=\frac{0.5}{2.0} \Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4} \Rightarrow \frac{v_{1}}{v_{2}}=\frac{1}{2} \)
Therefore, ratio of maximum speeds of emitted electrons will be \( 1: 2 \).
Hence, the correct answer is option: 'C'.