Question

Find the momentum of a photon if the wavelength of the radiation is 6630 angstroms.​

Answer 1

Answer:

The momentum of the incident radiation is given as p=  

λ

h

​  

.

When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of it's momentum without changing it's magnitude  

Δp=2p=  

λ

2h

​ =6630×10  

−10

 2×6.6×10  

−34

  =2×10  

−27

 kg−m/sec.

Explanation:

Hope it helps

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Answer 2

Answer:

According to De-Broglie's concept of wavelength,

$\boxed{ \lambda = \dfrac{h}{\rho}}$

where,

• λ = Wavelength of the particle
• h = Planck's constant
• ρ = Momentum of the particle

It is given that, λ = 6630 A°. Hence substituting in the formula we get:

$\implies \lambda = \dfrac{ 6.63 \times 10^{-34}}{\rho}\\\\\\\implies \rho = \dfrac{ 6.63 \times 10^{-34}}{ 6630 \times 10^{-10}}\\\\\\\implies \rho = \dfrac{6.63 \times 10^{-34}}{ 6.63 \times 10^{-7}}\\\\\\\implies \boxed{ \bf{\rho = 1 \times 10^{-34+7} = 10^{-27}\: Kg.m/s}}$

This is the required answer.