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  4. Photons of energy 1 eV and 2.5 eV successively illuminated a metal whose work function is 0.5 eV. The ratio of maximum speeds of emitted electron is

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Q. Photons of energy 1 eV and 2.5 eV successively illuminated a metal whose work function is 0.5 eV. The ratio of maximum speeds of emitted electron is

NTA AbhyasNTA Abhyas 2020

Solution:

Energy of photon $E=\phi+\frac{1}{2}mv^{2}$
$V_{max \, }=\sqrt{\frac{2 \left(E - \phi\right)}{m}}$
So putting values
$\frac{V_{max 1}}{V_{m a x \, 2} \, }=\sqrt{\frac{1 - 0.5}{2.5 - 0.5}}$
$\frac{V_{max 1}}{V_{max ⁡ 2}}=\frac{1}{2}$