The rest mass of an electron as well as that of positron is 0.51 MeV. When an electron and positron are annihilate, they produce gamma-rays of wavelength(s)

Options

A. 0.012 Å
B. 0.024 Å
C. `0.012 Å` to `infty`
D. `0.024 Å` to `infty`

Explanation

To solve the problem of finding the wavelength of gamma rays produced when an electron and a positron annihilate, we can follow these steps:

### Step 1: Calculate the total energy of the electron and positron
The rest mass energy of an electron (m_e) is given as 0.51 MeV. Since the positron has the same mass as the electron, its rest mass energy is also 0.51 MeV.

\[ E_{\text{total}} = E_{\text{electron}} + E_{\text{positron}} = 0.51 \, \text{MeV} + 0.51 \, \text{MeV} = 1.02 \, \text{MeV} \]

### Step 2: Convert the total energy from MeV to Joules
To convert MeV to Joules, we use the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\).

\[ E_{\text{total}} = 1.02 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 1.632 \times 10^{-13} \, \text{J} \]

### Step 3: Use the energy-wavelength relationship to find the wavelength
The relationship between energy (E) and wavelength (λ) is given by the equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))

Rearranging the equation to solve for wavelength gives:

\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the values into the equation
Now we can substitute the values of \(h\), \(c\), and \(E_{\text{total}}\) into the equation:

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{1.632 \times 10^{-13} \, \text{J}} \]

### Step 5: Calculate the wavelength
Calculating the above expression gives:

\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{1.632 \times 10^{-13} \, \text{J}} \approx 1.216 \times 10^{-12} \, \text{m} \]

### Step 6: Convert the wavelength to Angstroms
To convert meters to Angstroms (1 Angstrom = \(10^{-10}\) m):

\[ \lambda \approx 1.216 \times 10^{-12} \, \text{m} \times \frac{1 \, \text{Angstrom}}{10^{-10} \, \text{m}} \approx 0.01216 \, \text{Angstroms} \]

### Final Answer
The wavelength of the gamma rays produced when an electron and positron annihilate is approximately \(0.01216 \, \text{Angstroms}\).

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