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Planck's Quantum Hypothesis

A radiation o...

Question

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C=Velocity of light)

\frac{2 E}{C}

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\frac{2 E}{C^{2}}

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\frac{E}{C^{2}}

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\frac{E}{C}

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Solution

The correct option is D $\frac{2 E}{C}$
The radiation of energy is given by,

$E = \frac{h}{λ}$

Initial momentum of the radiation is

$P_{i} = - \frac{h}{λ} = - \frac{E}{C}$

The reflected momentum is

$P_{r} = - \frac{h}{λ} = - \frac{E}{C}$

So, the change in momentum of light is

$Δ P_{l i g h t} = P_{r} - P_{i} = - \frac{2 E}{C}$

Thus, the momentum transferred to the surface is,

$Δ P_{l i g h t} = P_{r} - P_{i} = - \frac{2 E}{C}$

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