Question

You scatter x-rays of wavelength 0.0105 nm from the loosely bound electrons of a target material. Find the angle at which the wavelength of the scattered x-rays is longer than the incident wavelength by a factor of =1.0367.

Answer 1

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