A nucleus kept at rest in free space, brakes up into smaller nuclei of masses `'m'` and `'2m'`. Total energy generated in this fission is `E`. The bigger part is radioactive, emits five gamma ray photons in the direction opposite to its velocity and finally comes torest. `["Given" h=6.6xx10^(-34)Js,m=1xx10^(-26)Kg,E= 3.63xx10^(-8) mc^(2), C=3xx10^(8)m//s)`
The wavelength of the gamma ray is

Options

A. `0.02Å`
B. `0.03Å`
C. `0.04Å`
D. `0.05Å`

Explanation

To find the wavelength of the gamma rays emitted during the radioactive decay of the larger nucleus, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.6 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^{8} \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the photon.

### Step 2: Calculate the total energy emitted by the gamma rays
The total energy generated in the fission is given as \(E = 3.63 \times 10^{-8} \, mc^2\). Since the larger nucleus emits 5 gamma photons, the energy of each gamma photon can be calculated as:
\[ E_{\text{photon}} = \frac{E}{5} \]

### Step 3: Substitute the values into the energy equation
We can rearrange the energy equation to solve for the wavelength \(\lambda\):
\[ \lambda = \frac{hc}{E_{\text{photon}}} \]

### Step 4: Substitute \(E_{\text{photon}}\) into the equation
Substituting \(E_{\text{photon}} = \frac{E}{5}\) into the wavelength equation gives:
\[ \lambda = \frac{hc}{\frac{E}{5}} = \frac{5hc}{E} \]

### Step 5: Substitute known values
Now, substituting the known values:
- \(h = 6.6 \times 10^{-34} \, \text{Js}\)
- \(c = 3 \times 10^{8} \, \text{m/s}\)
- \(E = 3.63 \times 10^{-8} \, mc^2\)

We can directly substitute \(E\) into the equation:
\[ \lambda = \frac{5 \times (6.6 \times 10^{-34}) \times (3 \times 10^{8})}{3.63 \times 10^{-8} \, mc^2} \]

### Step 6: Simplify the expression
To simplify, we can calculate \(mc^2\):
\[ mc^2 = (1 \times 10^{-26} \, \text{kg}) \times (3 \times 10^{8} \, \text{m/s})^2 = 9 \times 10^{-9} \, \text{J} \]
Now substituting this value into the equation for \(\lambda\):
\[ E = 3.63 \times 10^{-8} \times (9 \times 10^{-9}) = 3.27 \times 10^{-16} \, \text{J} \]

### Step 7: Calculate the wavelength
Now substituting back into the wavelength equation:
\[ \lambda = \frac{5 \times (6.6 \times 10^{-34}) \times (3 \times 10^{8})}{3.27 \times 10^{-16}} \]
Calculating this gives:
\[ \lambda \approx 2 \times 10^{-12} \, \text{m} \]

### Step 8: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \(10^{-10}\) m):
\[ \lambda = 2 \times 10^{-12} \, \text{m} = 0.02 \, \text{Å} \]

### Final Answer
The wavelength of the gamma rays emitted is:
\[ \lambda = 0.02 \, \text{Å} \]