Question

The de Broglie wavelength of a proton is twice the de Broglie wavelength of an alpha particle. The ratio of the kinetic energies of the proton and the alpha particle is:

• A. \( 1:1 \)
• B. \( 1:4 \)
• C. \( 1:2 \)
• D. \( 1:8 \) \bigskip

Hint:

The de Broglie wavelength \( \lambda \) is inversely proportional to the momentum of the particle. The kinetic energy \( K \) is proportional to the square of the momentum. By comparing the kinetic energies, we can determine the relationship between the masses and momenta of the proton and alpha particle.

Answer 1

The Correct Option is A

We are given that the de Broglie wavelength \( \lambda \) of a proton is twice that of an alpha particle, i.e., \[ \lambda_{\text{proton}} = 2 \lambda_{\text{alpha}} \] The de Broglie wavelength \( \lambda \) of a particle is related to its momentum \( p \) by the formula: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. 

Step 1: Let the momentum of the proton be \( p_{\text{proton}} \) and the momentum of the alpha particle be \( p_{\text{alpha}} \). From the de Broglie relation, we have: \[ \lambda_{\text{proton}} = \frac{h}{p_{\text{proton}}}, \quad \lambda_{\text{alpha}} = \frac{h}{p_{\text{alpha}}} \] Since \( \lambda_{\text{proton}} = 2 \lambda_{\text{alpha}} \), we can write: \[ \frac{h}{p_{\text{proton}}} = 2 \times \frac{h}{p_{\text{alpha}}} \] This simplifies to: \[ p_{\text{proton}} = \frac{p_{\text{alpha}}}{2} \] 

Step 2: The kinetic energy \( K \) of a particle is related to its momentum \( p \) by the equation: \[ K = \frac{p^2}{2m} \] where \( m \) is the mass of the particle.

 Step 3: Let the masses of the proton and alpha particle be \( m_{\text{proton}} \) and \( m_{\text{alpha}} \), respectively. Then the kinetic energy of the proton \( K_{\text{proton}} \) and the alpha particle \( K_{\text{alpha}} \) are given by: \[ K_{\text{proton}} = \frac{p_{\text{proton}}^2}{2 m_{\text{proton}}}, \quad K_{\text{alpha}} = \frac{p_{\text{alpha}}^2}{2 m_{\text{alpha}}} \] Substituting \( p_{\text{proton}} = \frac{p_{\text{alpha}}}{2} \) into the equation for \( K_{\text{proton}} \), we get: \[ K_{\text{proton}} = \frac{\left(\frac{p_{\text{alpha}}}{2}\right)^2}{2 m_{\text{proton}}} = \frac{p_{\text{alpha}}^2}{8 m_{\text{proton}}} \] 

Step 4: The ratio of the kinetic energies is: \[ \frac{K_{\text{proton}}}{K_{\text{alpha}}} = \frac{\frac{p_{\text{alpha}}^2}{8 m_{\text{proton}}}}{\frac{p_{\text{alpha}}^2}{2 m_{\text{alpha}}}} = \frac{1}{4} \times \frac{m_{\text{alpha}}}{m_{\text{proton}}} \] Since the mass of the alpha particle is 4 times the mass of the proton (as it consists of 2 protons and 2 neutrons), we have: \[ \frac{K_{\text{proton}}}{K_{\text{alpha}}} = \frac{1}{4} \times \frac{4}{1} = 1 \] Thus, the ratio of the kinetic energies of the proton and the alpha particle is \( 1:1 \).  

Conclusion: Therefore, the correct answer is \( \boxed{1:1} \).