Respuesta :
Answer:
To solve this problem, we can use the relativistic kinetic energy formula:
\[ K = (\gamma - 1)mc^2 \]
where:
- \( K \) is the kinetic energy,
- \( \gamma \) is the Lorentz factor,
- \( m \) is the rest mass of the particle, and
- \( c \) is the speed of light.
We know the kinetic energy \( K = 4.75 \) MeV and the rest mass \( m = 6.64 \times 10^{-27} \) kg. We need to find the Lorentz factor \( \gamma \) to determine the velocity of the alpha particle relative to the speed of light.
First, let's convert the kinetic energy from MeV to joules:
\[ 1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J} \]
So, \( 4.75 \, \text{MeV} = 4.75 \times 1.602 \times 10^{-13} \, \text{J} = 7.5975 \times 10^{-13} \, \text{J} \).
Now, we can rearrange the relativistic kinetic energy formula to solve for \( \gamma \):
\[ \gamma = \frac{K}{mc^2} + 1 \]
Substituting the given values:
\[ \gamma = \frac{7.5975 \times 10^{-13} \, \text{J}}{(6.64 \times 10^{-27} \, \text{kg})(3.00 \times 10^8 \, \text{m/s})^2} + 1 \]
\[ \gamma = \frac{7.5975 \times 10^{-13}}{(6.64 \times 10^{-27})(9 \times 10^{16})} + 1 \]
\[ \gamma = \frac{7.5975 \times 10^{-13}}{5.976 \times 10^{-10}} + 1 \]
\[ \gamma ≈ 1.269 \]
Now, the velocity of the alpha particle as a ratio to the speed of light (\( \frac{v}{c} \)) can be found using the formula for the Lorentz factor:
\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Solving for \( \frac{v}{c} \):
\[ \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}} \]
\[ \frac{v}{c} = \sqrt{1 - \frac{1}{(1.269)^2}} \]
\[ \frac{v}{c} ≈ \sqrt{1 - 0.625} \]
\[ \frac{v}{c} ≈ \sqrt{0.375} \]
\[ \frac{v}{c} ≈ 0.612 \]
So, the velocity of the alpha particle is approximately 0.612 times the speed of light.
