Two radiations of photons energies $1 e V$ and $2.5 e V$ successively illuminate a photosensitive metallic surface of work function $0.5 e V$ . The ratio of the maximum speeds of the emitted electrons is-

1 : 2

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1 : 1

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1 : 5

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1 : 4

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Solution

The correct option is A $1 : 2$
We know that,
$v_{m a x} = \sqrt{\frac{2}{m} (K E_{m a x})} = \sqrt{\frac{2}{m} (E_{P h} - W)}$

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{1 - 0.5}{2.5 - 0.5}} = \sqrt{\frac{0.5}{2}} = 1 : 2$

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Two radiations of photons energies 1eV and 2.5eV successively illuminate a photosensitive metallic surface of work function 0.5eV. The ratio of the maximum speeds of the emitted electrons is-

The correct option is A 1:2We know that, vmax=√2m(KEmax)=√2m(EPh−W)v1v2=√1−0.52.5−0.5=√0.52=1:2

Two radiations of photons energies $1 e V$ and $2.5 e V$ successively illuminate a photosensitive metallic surface of work function $0.5 e V$ . The ratio of the maximum speeds of the emitted electrons is-

The correct option is A $1 : 2$
We know that,
$v_{m a x} = \sqrt{\frac{2}{m} (K E_{m a x})} = \sqrt{\frac{2}{m} (E_{P h} - W)}$

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{1 - 0.5}{2.5 - 0.5}} = \sqrt{\frac{0.5}{2}} = 1 : 2$