An X-rays of wavelength 0.140 nm are scattered from a block of carbon. What will be the wavelengths of X-rays scattered at `90^(@)`?

Options

A. 01.40 nm
B. 0.142 nm
C. 01.44 nm
D. 01.46 nm

Explanation

To solve the problem of finding the wavelength of X-rays scattered at an angle of \(90^\circ\), we can use the Compton scattering formula. Here’s a step-by-step solution:

### Step 1: Understand the Compton Scattering Formula
The change in wavelength (\(\Delta \lambda\)) due to Compton scattering is given by the formula:
\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \phi) \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
- \(m_e\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))
- \(\phi\) is the scattering angle.

### Step 2: Substitute the Values
Since the scattering angle \(\phi\) is \(90^\circ\), we have:
\[ \cos 90^\circ = 0 \]
Thus, the formula simplifies to:
\[ \Delta \lambda = \frac{h}{m_e c} (1 - 0) = \frac{h}{m_e c} \]

### Step 3: Calculate \(\Delta \lambda\)
Now, we can substitute the known values into the equation:
\[ \Delta \lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 3 \times 10^8} \]
Calculating the denominator:
\[ 9.1 \times 10^{-31} \times 3 \times 10^8 = 2.73 \times 10^{-22} \]
Now substituting back:
\[ \Delta \lambda = \frac{6.626 \times 10^{-34}}{2.73 \times 10^{-22}} \approx 2.43 \times 10^{-12} \, \text{m} = 0.00243 \, \text{nm} \]

### Step 4: Find the Final Wavelength
The final wavelength (\(\lambda_f\)) after scattering can be calculated as:
\[ \lambda_f = \lambda_i + \Delta \lambda \]
where \(\lambda_i\) is the initial wavelength (\(0.140 \, \text{nm}\)):
\[ \lambda_f = 0.140 \, \text{nm} + 0.00243 \, \text{nm} = 0.14243 \, \text{nm} \]
Rounding this gives:
\[ \lambda_f \approx 0.142 \, \text{nm} \]

### Final Answer
The wavelength of X-rays scattered at \(90^\circ\) is approximately \(0.142 \, \text{nm}\).

---