JIf the photon scatters at an angle of 60.0∘ from its original direction, what is the magnitude of the linear momentum of the electron just after it collides with the photon?

(7.558 × 10⁻²⁶) kgm/s

Explanation:

A diagram of the collision is presented in the attached image.

Let the wavelength of the photon before collision = λ

Wavelength of the photon after collision = λ'

Frequency of the photon before collision = f

Frequency of the photon after collision = f'

The momentum of the photon before collision, Pₚ = (h/λ)

The momentum of the photon before collision, Pₚ' = (h/λ')

where h = Planck's constant

Let the momentum of the electron after collision be P

Compton theorizes that the elastic collision of the photon and electron obeys the laws of conservation of energy and momentum.

The conservation of Momentum

Momentum before collision = momentum after collision

Momentum before collision = Pₚ

Momentum after collision = (Pₚ' + P)

Pₚ = (Pₚ' + P)

P = (Pₚ - Pₚ') = (h/λ) - (h/λ') = h[(1/λ) - (1/λ')]

All of these variables to find the momentum of the electron after collision are known apart from the wavelength of the photon after collision.

To find this, we use the Compton's wavelength formula

Starting from the conservation of mass and momentum, Compton proved that the wavelength of the electron after collision, λ' is given by

λ' - λ = (h/m₀c) (1 - cos θ)

where

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

m₀ = mass of the electron = 9.11 × 10⁻³¹ kg

c = speed of light = 3.0 × 10⁸ m/s

θ = angle of scattering of the photons after collision = 60°; cos 60° = 0.5

λ = 0.102 nm = 1.02 × 10⁻¹⁰ m

λ' - λ = [(6.63 × 10⁻³⁴/(9.11 × 10⁻³¹ × 3.0 × 10⁸)] × 0.5

λ' - λ = 1.213 × 10⁻¹² m

λ' = λ + (1.213 × 10⁻¹²) = (1.02 × 10⁻¹⁰) + (1.213 × 10⁻¹²) = (1.032 × 10⁻¹²) m

P = (Pₚ - Pₚ') = (h/λ) - (h/λ') = h[(1/λ) - (1/λ')]

P = h[(1/λ) - (1/λ')] = (6.63 × 10⁻³⁴) [(1/(1.02 × 10⁻¹⁰)) - (1/(1.032 × 10⁻¹²))] = (6.63 × 10⁻³⁴) (113999088) = (7.558 × 10⁻²⁶) kgm/s

think its the third option

the correct choice is a .

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