To solve the problem step by step, we will follow the outlined procedure to find the kinetic energy of the nucleus after emitting a gamma photon.
### Step 1: Convert the mass of the nucleus from atomic mass units (u) to kilograms.
Given:
- Mass of the nucleus = 20 u
- Conversion factor: \( 1 \, \text{u} = 1.6 \times 10^{-27} \, \text{kg} \)
Calculation:
\[
\text{Mass in kg} = 20 \, \text{u} \times 1.6 \times 10^{-27} \, \text{kg/u} = 32 \times 10^{-27} \, \text{kg} = 3.2 \times 10^{-26} \, \text{kg}
\]
### Step 2: Convert the energy of the gamma photon from MeV to Joules.
Given:
- Energy of the photon = 6 MeV
- Conversion factor: \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \)
Calculation:
\[
\text{Energy in Joules} = 6 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 9.6 \times 10^{-13} \, \text{J}
\]
### Step 3: Calculate the momentum of the emitted gamma photon.
Using the formula for momentum \( p \):
\[
p = \frac{E}{c}
\]
where \( c \) (speed of light) is approximately \( 3 \times 10^8 \, \text{m/s} \).
Calculation:
\[
p = \frac{9.6 \times 10^{-13} \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 3.2 \times 10^{-21} \, \text{kg m/s}
\]
### Step 4: Calculate the kinetic energy of the nucleus using the momentum.
The kinetic energy \( KE \) can be expressed in terms of momentum:
\[
KE = \frac{p^2}{2m}
\]
Substituting the values:
\[
KE = \frac{(3.2 \times 10^{-21})^2}{2 \times (3.2 \times 10^{-26})}
\]
Calculation:
\[
KE = \frac{10.24 \times 10^{-42}}{6.4 \times 10^{-26}} = 1.6 \times 10^{-16} \, \text{J}
\]
### Step 5: Convert the kinetic energy from Joules to electron volts.
Using the conversion \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
KE \text{ (in eV)} = \frac{1.6 \times 10^{-16} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 1000 \, \text{eV} = 1 \, \text{keV}
\]
### Final Answer:
The kinetic energy of the nucleus after emitting the gamma photon is approximately **1 keV**.
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