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  4. Light of two different frequencies whose photons have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function 0.5eV successively. The ratio of maximum speeds of emitted electrons will be

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Q. Light of two different frequencies whose photons have energies $1eV$ and $2.5eV$ respectively illuminate a metallic surface whose work function $0.5eV$ successively. The ratio of maximum speeds of emitted electrons will be

NTA AbhyasNTA Abhyas 2022

Solution:

Here work function, $\phi_{0}=0.5eV$
According to Einstein's photoelectric equation
$Maximumenergyofemittedelectron=Incidentphotonenergy-Workfunction$
∴ $K_{\left(max\right)_{1}}=1.eV-0.5eV=0.5eV...\left(i\right)$
and $K_{\left(max\right)_{2}}=2.5eV-0.5eV=2eV...\left(ii\right)$
On dividing equation $\left(i\right)$ by equation $\left(ii\right)$ , we get
$\frac{K_{max_{1}}}{K_{max_{2}}}=\frac{0 \cdot 5 e V}{2 e V}=\frac{1}{4}$
$\Rightarrow \frac{\frac{1}{2} m v_{max_{1}}^{2}}{\frac{1}{2} m v_{max_{2}}^{2}}=\frac{1}{4}$
$\Rightarrow \frac{v_{max_{1}}}{v_{max_{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$