Question

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

Answer 1

Initial momentum of surface, pi=Ec

Since surface is perfectly reflecting, so, final momentum, pf=E/c

So, change in momentum, Δp=pf−pi=−2E/c

So, momentum transferred to the surface is Δp′=|Δp|=2Ec

Answer 2

Theory :-

From Max Planck's equation of photoelectric effect,

Momentum of a striking photon,

$p=\frac{h}{\lambda},\ where\ h=Planck's\ constant;\lambda=Wavelength\ of\ striking\ photon$

Also, we know that energy of a moving photon particle, $E=\frac{hc}{\lambda},\ where\ c=Speed\ of\ light$

So, comparing both the equations, we get

$p=\frac{E}{c}$

Solution :-

Initial momentum of photon

$=p_i=\frac{E}{c}(\hat{i})$

Since, finally photon collides elastically to the surface,

Final momentum of photon

$=p_f=\frac{E}{c}(-\hat{i})$

Hence, this change in momentum is imparted to the photon

And, negative of this change is imparted to the surface.

Hence, momentum transferred to the surface is (Only magnitude)

$=\frac{E}{c}-(-\frac{E}{c})\\=2\frac{E}{c}$

Hence, correct answer is 2E/c

Remarks :-

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