Step 1: Kinetic energy of proton (Non-relativistic)
$KE_p = \frac{P_1^2}{2m_0} = \frac{1}{8} m_0 c^2$
Step 2: Solve for $P_1$
$P_1^2 = \frac{2m_0}{8} m_0 c^2 = \frac{1}{4} m_0^2 c^2$
$P_1 = \sqrt{\frac{1}{4} m_0^2 c^2} = \frac{1}{2} m_0 c$
Step 3: Momentum of photon ($P_2$)
Energy of photon $E_{photon} = KE_p = \frac{1}{8} m_0 c^2$
Momentum of photon $P_2 = \frac{E_{photon}}{c} = \frac{\frac{1}{8} m_0 c^2}{c} = \frac{1}{8} m_0 c$
Step 4: Calculate $\frac{P_1 - P_2}{P_1}$
$\frac{P_1 - P_2}{P_1} = 1 - \frac{P_2}{P_1} = 1 - \frac{\frac{1}{8} m_0 c}{\frac{1}{2} m_0 c} = 1 - \frac{1/8}{1/2} = 1 - \frac{2}{8} = 1 - \frac{1}{4} = \frac{3}{4}$
Step 5: Match to Option
Option (D) is $\frac{3}{4}$.
Final Answer: The final answer is $\boxed{\frac{3}{4}}$
Given:
Step 1: Express proton's kinetic energy
Given \( K = \frac{1}{8}m_pc^2 \), where \( m_p \) is proton rest mass.
Step 2: Relate kinetic energy to momentum (relativistically)
Total energy \( E = K + m_pc^2 = \frac{9}{8}m_pc^2 \)
Using energy-momentum relation:
\[ E^2 = (P_1c)^2 + (m_pc^2)^2 \]
\[ \left(\frac{9}{8}m_pc^2\right)^2 = (P_1c)^2 + (m_pc^2)^2 \]
\[ \frac{81}{64}m_p^2c^4 = P_1^2c^2 + m_p^2c^4 \]
\[ P_1^2c^2 = \frac{17}{64}m_p^2c^4 \]
\[ P_1 = \frac{\sqrt{17}}{8}m_pc \]
Step 3: Calculate photon momentum \( P_2 \)
Photon energy \( E_\gamma = K = \frac{1}{8}m_pc^2 \)
Photon momentum:
\[ P_2 = \frac{E_\gamma}{c} = \frac{m_pc}{8} \]
Step 4: Compute the ratio \( \frac{P_1 - P_2}{P_1} \)
\[ \frac{P_1 - P_2}{P_1} = 1 - \frac{P_2}{P_1} \]
\[ = 1 - \frac{\frac{m_pc}{8}}{\frac{\sqrt{17}}{8}m_pc} \]
\[ = 1 - \frac{1}{\sqrt{17}} \]
\[ ≈ 1 - 0.2425 ≈ 0.7575 \]
An object has moved through a distance can it have zero displacement if yes support your answer with an example.