Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5 eV. The ratio of maximum kinetic energy of the emitted electron will
be

1 : 5

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1 : 4

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1 : 2

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1 : 1

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Solution

The correct option is B 1 : 4
By using $K_{m a x} = E - W_{0} \Rightarrow \frac{(K_{m a x})_{1}}{(K_{m a x})_{2}} = \frac{1 - 0.5}{2.5 - 0.5} = \frac{0.5}{2} = \frac{1}{4}$ .

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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5 eV. The ratio of maximum kinetic energy of the emitted electron will be

The correct option is B 1 : 4By using Kmax=E−W0⇒(Kmax)1(Kmax)2=1−0.52.5−0.5=0.52=14.

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5 eV. The ratio of maximum kinetic energy of the emitted electron will
be

The correct option is B 1 : 4
By using $K_{m a x} = E - W_{0} \Rightarrow \frac{(K_{m a x})_{1}}{(K_{m a x})_{2}} = \frac{1 - 0.5}{2.5 - 0.5} = \frac{0.5}{2} = \frac{1}{4}$ .