Problem 21 - University Physics with Modern Physics

Question

X rays with an initial wavelength of 0.900 \(\times\) 10\(^{-10}\) m undergo Compton scattering. For what scattering angle is the wavelength of the scattered x rays greater by 1.0\(\%\) than that of the incident x rays?

Step-by-Step Solution

Answer

The scattering angle is approximately 51.14°.

Step 1: Identify Given Quantities

We are given the initial wavelength of the X rays as \( \lambda_i = 0.900 \times 10^{-10} \ \text{m} \) and the desired increase in wavelength of 1%. This means the final wavelength \( \lambda_f \) is \( \lambda_f = \lambda_i + 0.01\lambda_i = 1.01\lambda_i \).

Step 2: Compton Wavelength Shift Formula

The change in wavelength due to Compton scattering is given by the formula: \[ \Delta \lambda = \lambda_f - \lambda_i = \frac{h}{m_e c} (1 - \cos \theta) \] where \( h \) is Planck's constant, \( m_e \) is the electron mass, \( c \) is the speed of light, and \( \theta \) is the scattering angle we need to find.

Step 3: Set Up the Equation

Substitute \( \Delta \lambda \) and \( \lambda_f \) in the formula: \[ 0.01 \times 0.900 \times 10^{-10} = \frac{h}{m_e c} (1 - \cos \theta) \].

Step 4: Calculate Compton Wavelength Constant

Calculate the Compton wavelength constant \( \frac{h}{m_e c} \), which is approximately \( 2.43 \times 10^{-12} \ \text{m}. \)

Step 5: Solve for Scattering Angle

Rearrange the equation to solve for \( \theta \): \[ 1 - \cos \theta = \frac{0.01 \times 0.900 \times 10^{-10}}{2.43 \times 10^{-12}} \], then solve \( \cos \theta = 1 - \frac{0.01 \times 0.900 \times 10^{-10}}{2.43 \times 10^{-12}} \).

Step 6: Determine Scattering Angle

Calculate \( \cos \theta \) to find the scattering angle \( \theta \) using a calculator: \[ \theta = \cos^{-1}(0.629) \], which gives \( \theta \approx 51.14^\circ \).

Key Concepts

Wavelength Shift in Compton Scattering

In Compton scattering, X rays collide with electrons and their path is altered, leading to a shift in their wavelength. This happens because when X rays interact with an electron, part of their energy is transferred to the electron. The result is a decrease in the energy of the X rays, causing a longer wavelength. This change in wavelength is called the "wavelength shift." For our exercise, the initial wavelength is given as \( \lambda_i = 0.900 \times 10^{-10} \ \text{m} \).The problem states that the wavelength of the scattered X rays will be 1% longer than the incident X rays' wavelength. So, the final wavelength \( \lambda_f \) becomes:

• \( \lambda_f = 1.01 \lambda_i \)

To quantify the wavelength shift \( \Delta \lambda \), we use the formula:

• \( \Delta \lambda = \lambda_f - \lambda_i \)
• \( \Delta \lambda = 0.01 \times 0.900 \times 10^{-10} \ \text{m} \)

This shift directly depends on the scattering angle, which becomes our next crucial concept.

The Role of Scattering Angle

The scattering angle \( \theta \) is essential in understanding how X rays are deflected when they hit an electron. It is the angle between the direction of the incoming and outgoing beam of X rays. The bigger the scattering angle, the more the X rays' wavelength is shifted. This relationship is described by the Compton scattering formula:\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]In this equation:

• \( \Delta \lambda \) is the change in wavelength.
• \( h \) is Planck's constant.
• \( m_e \) is the electron's mass.
• \( c \) is the speed of light.
• \( \cos \theta \) quantifies the angle of scattering.

By rearranging the formula, we can solve for the scattering angle \( \theta \) when the wavelength shift is known. For this exercise:

• Calculate \( 1 - \cos \theta = \frac{\Delta \lambda}{\frac{h}{m_e c}} \)
• Thus, \( \theta = \cos^{-1}(0.629) \)
• This makes \( \theta \approx 51.14^\circ \), showing a significant deviation.

Comprehending the scattering angle aids in visualizing how the interaction and deflection happen.

Understanding the Compton Wavelength

The Compton wavelength is another key element in understanding the scattering of X rays. It is a fundamental constant derived from Planck's constant, the electron's mass, and the speed of light. It represents the wavelength shift when the scattering angle is \( 90^\circ \). Generally:\[ \lambda_c = \frac{h}{m_e c} \]This constant's value is approximately \( 2.43 \times 10^{-12} \ \text{m} \). The Compton wavelength helps us predict the extent of the wavelength shift in Compton scattering. It is vital for solving equations where the scattering angle and wavelength shift are related, such as in the initial exercise:\[ 1 - \cos \theta = \frac{0.01 \times 0.900 \times 10^{-10}}{2.43 \times 10^{-12}} \]The Compton wavelength constantly bridges fundamental physical constants to provide insights into particle interactions. It illustrates the quantifiable aspects of wave-particle duality that arise in quantum mechanics and scattering processes.