x ray of wavelength of 10 pm are scattered from target scattered wavlength is 10.7 pm

find maximum wavelength

this is based on compton effect

To tackle this problem, we need to apply the principles of the Compton effect, which describes how X-rays scatter off electrons. The key concept here is that when X-rays collide with electrons, they can transfer some of their energy to the electrons, resulting in a change in the wavelength of the scattered X-rays. This change in wavelength can be calculated using the Compton wavelength shift equation.

Understanding the Compton Wavelength Shift

The Compton effect is characterized by the equation:

Δλ = λ' - λ = \frac{h}{m_e c} (1 - \cos \theta)

Where:

• Δλ is the change in wavelength.
• λ' is the wavelength of the scattered X-ray.
• λ is the initial wavelength of the X-ray.
• h is Planck's constant (approximately 6.626 x 10^-34 Js).
• m_e is the mass of the electron (approximately 9.11 x 10^-31 kg).
• c is the speed of light (approximately 3.00 x 10⁸ m/s).
• θ is the scattering angle.

Given Values

From your question, we have:

• Initial wavelength (λ) = 10 pm = 10 x 10^-12 m
• Scattered wavelength (λ') = 10.7 pm = 10.7 x 10^-12 m

Calculating the Change in Wavelength

First, we can find the change in wavelength (Δλ):

Δλ = λ' - λ = (10.7 x 10^-12 m) - (10 x 10^-12 m) = 0.7 x 10^-12 m = 0.7 pm

Finding the Maximum Wavelength

The maximum wavelength occurs when the scattering angle θ is 180 degrees (which means the X-ray is scattered directly backward). In this case, the cosine term becomes -1:

Δλ_max = \frac{h}{m_e c} (1 - (-1)) = \frac{2h}{m_e c}

Now, we can calculate the maximum change in wavelength:

Δλ_max = 2 \cdot \frac{(6.626 \times 10^{-34} \text{ Js})}{(9.11 \times 10^{-31} \text{ kg}) \cdot (3.00 \times 10^{8} \text{ m/s})}

Calculating this gives:

Δλ_max ≈ 4.86 \times 10^{-12} \text{ m} = 4.86 \text{ pm}

Final Calculation of Maximum Wavelength

Now, we can find the maximum wavelength (λ_max) by adding this maximum change to the initial wavelength:

λ_max = λ + Δλ_max = (10 \times 10^{-12} \text{ m}) + (4.86 \times 10^{-12} \text{ m}) = 14.86 \times 10^{-12} \text{ m} = 14.86 \text{ pm}

Summary

In summary, the maximum wavelength of the scattered X-ray when using the Compton effect with an initial wavelength of 10 pm is approximately 14.86 pm. This illustrates how the energy transfer during scattering can significantly alter the properties of X-rays, which is a fundamental concept in quantum physics and medical imaging technologies.