Answer:
approximately 0.000271 degrees.
Step-by-step explanation:
To find the angle at which the wavelength of the scattered x-rays is longer than the incident wavelength by a factor of 1.0367, we can use the Compton scattering formula:
Δλ = λ' - λ = h / (m_ec) * (1 - cosθ)
where:
Δλ is the change in wavelength,
λ' is the scattered wavelength,
λ is the incident wavelength,
h is the Planck's constant (6.626 x 10^(-34) J·s),
m_e is the electron's rest mass (9.109 x 10^(-31) kg),
c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s),
θ is the scattering angle.
We are given:
λ = 0.0105 nm = 0.0105 x 10^(-9) m
Factor of change (λ'/λ) = 1.0367
We can rewrite the factor of change as:
1 + Δλ/λ = λ'/λ
Now, solve for Δλ:
Δλ = (λ'/λ - 1) * λ
Substitute the given values:
Δλ = (1.0367 - 1) * 0.0105 x 10^(-9) m
Δλ = 0.0367 * 0.0105 x 10^(-9) m
Δλ ≈ 3.8535 x 10^(-11) m
Now, we can find the scattering angle (θ) using the Compton scattering formula:
Δλ = h / (m_ec) * (1 - cosθ)
Rearrange the formula to solve for cosθ:
cosθ = 1 - (Δλ * m_ec) / h
Substitute the known values:
cosθ = 1 - (3.8535 x 10^(-11) m * 9.109 x 10^(-31) kg * 3.00 x 10^8 m/s) / (6.626 x 10^(-34) J·s)
Now, calculate cosθ and then find θ:
cosθ ≈ 1 - 1.0308 x 10^(-11)
cosθ ≈ 0.999999999989692
θ ≈ arccos(0.999999999989692)
θ ≈ 4.738 x 10^(-6) radians
Finally, convert the scattering angle to degrees:
θ ≈ 4.738 x 10^(-6) radians * (180° / π radians)
θ ≈ 0.000271 degrees