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VITEEE 2011 Paper

Section: Physics

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Question : 19 of 120

Marks: +1, -0

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speed of the emitted electrons will be

1 : 5
1 : 4
1 : 2
1 : 1

Solution:

For maximum speed of the photo electrons

\frac{1}{2} mv^{2} = E_{\text{photon}} - \phi

where

\phi = 0.5 \mathrm{eV}

is the work function of the metal.

Energy of photon

E_{1} = 1 \mathrm{eV}

\therefore \frac{1}{2} mv_{1}^{2} = 1 - 0.5

We get

\frac{1}{2} mv_{1}^{2} = 0.5 \mathrm{eV}

....(1)

Energy of photon

E_{2} = 2.5 \mathrm{eV}

\therefore \frac{1}{2} mv_{2}^{2} = 2.5 - 0.5

We get

\frac{1}{2} mv_{2}^{2} = 2 \mathrm{eV}

....(2)

Ratio of maximum speeds

\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{0.5}{2} = \frac{1}{4}

We get

\frac{v_{1}}{v_{2}} = \frac{1}{2}

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