Light of two different frequencies whose photons have energies 1eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5eV. 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 $E = W_{0} + K_{m a x} \Rightarrow K_{m a x} = E - W_{0}$
$H e n c e, K_{1} = 1 - 0.5 = 0.5$
$a n d K_{2} = 2.5 - 0.5 = 2 \Rightarrow \frac{K_{1}}{K_{2}} = \frac{1}{4}$

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

The correct option is B 1 : 4By using E=W0+Kmax⇒Kmax=E−W0 Hence, K1=1−0.5=0.5 and K2=2.5−0.5=2⇒K1K2=14

The correct option is B 1 : 4
By using $E = W_{0} + K_{m a x} \Rightarrow K_{m a x} = E - W_{0}$
$H e n c e, K_{1} = 1 - 0.5 = 0.5$
$a n d K_{2} = 2.5 - 0.5 = 2 \Rightarrow \frac{K_{1}}{K_{2}} = \frac{1}{4}$