Derive the Compton wavelength shift formula and then find the wavelength of a photon that is initially $0.05 nm$ and scattered elastically by a stationary electron at an angle of $6 0^{\circ}$ .

hard 5 marks Class 12 · Physics · Dual Nature of Radiation and Matter

Answer

λ^{'} = 5.12 \times 1 0^{- 2} nm

The correct answer is $λ^{'} = 5.12 \times 1 0^{- 2} nm$ — Rounded to three significant figures, $λ^{'} = 5.12 \times 1 0^{- 2} nm$ .

Step-by-step solution

Consider a photon of initial wavelength $λ$ colliding with a stationary electron; after scattering at angle $θ$ , its wavelength becomes $λ^{'}$ .
Conserve energy: $\frac{h c}{λ} + m_{e} c^{2} = \frac{h c}{λ ^{'}} + (p_{e} c)^{2} + m_{e}^{2} c^{4}$ , where $p_{e}$ is the electron momentum after collision.
Conserve momentum in the direction of the photon and perpendicular to it, leading to the relation $\frac{h}{λ} - \frac{h}{λ ^{'}} = p_{e} cos ϕ$ and $p_{e} sin ϕ = \frac{h}{λ ^{'}} sin θ$ .
Eliminate $p_{e}$ and $ϕ$ to obtain $λ^{'} - λ = \frac{h}{m _{e} c} (1 - cos θ)$ , the Compton shift formula.
The constant $\frac{h}{m _{e} c} = λ_{c} = 2.426 \times 1 0^{- 12} m = 0.002426 nm$ (Compton wavelength of electron).
For $θ = 6 0^{\circ}$ , $1 - cos θ = 1 - 0.5 = 0.5$ .
Hence $Δ λ = λ_{c} \times 0.5 = 0.002426 \times 0.5 = 0.001213 nm$ .
The scattered wavelength is $λ^{'} = λ + Δ λ = 0.05 + 0.001213 = 0.051213 nm$ .
Rounded to three significant figures, $λ^{'} = 5.12 \times 1 0^{- 2} nm$ .

Common mistake: Using the electron rest mass energy instead of the Compton wavelength constant or mixing up initial and scattered wavelengths.

Concept tested: Compton effect – wavelength shift derivation and calculation