Question

In Compton collision, the incident radiation of wavelength

1.087Å is scattered from a scatterer at an angle of 300

then

the wavelength of scattered photon is given by,​

Answer 1

Answer:

A photon of wavelength 6000 nm collides with an electron at rest. After scattering, the wavelength of the scattered photon is found to change by exactly one Compton wavelength.

Answer 2

Answer:

The wavelength of the scattered photon is 1.09025A°.

Explanation:

We will use the Crompton relation to solve this question. Which is given as,

$\lambda_2-\lambda_1=\frac{h}{m_ec}(1-cos\theta)$        (1)

Where,

λ₂=wavelength of the scattered photon

λ₁=wavelength of the incident photon

h=plancks constant=6.626×10⁻³⁴ Joule-second

$m_e$=mass of an electron=9.1×10⁻³¹kg

c=speed of light=3×10⁸m/s

θ=angle of scattering

From the question we have,

λ₁=1.087A°

θ=30°

By substituting the required values in equation (1) we get;

$\lambda_2-1.087=\frac{6.626\times 10^-^3^4}{9.1\times 10^-^3^1\times 3\times 10^8}(1-cos30\textdegree)$     (2)

cos30°=0.866

By placing the value of cos30° in equation (2) we get;

$\lambda_2-1.087=\frac{6.626\times 10^-^3^4}{9.1\times 10^-^3^1\times 3\times 10^8}(1-0.866)$

$\lambda_2-1.087=\frac{6.626\times 10^-^3^4}{9.1\times 10^-^3^1\times 3\times 10^8}\times 0.134$

$\lambda_2-1.087A\textdegree=0.00325A\textdegree$

$\lambda_2=0.00325A\textdegree+1.087A\textdegree$

$\lambda_2=1.09025A\textdegree$

Hence, the wavelength of the scattered photon is 1.09025A°.