Question

The de Broglie wavelength of a proton and \( \alpha \)-particle are equal. The ratio of their velocities is:

• A. 4 : 3
• B. 4 : 1
• C. 4 : 2
• D. 1 : 4

Hint:

The de Broglie wavelength is inversely proportional to both the mass and velocity of the particle. For equal wavelengths, the ratio of velocities is the inverse ratio of the square roots of the masses.

Answer 1

The Correct Option is B

Step 1: The de Broglie wavelength \( \lambda \) is given by the formula: \[ \lambda = \frac{h}{mv}, \] where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is the velocity. 
Step 2: Since the de Broglie wavelength of the proton and the \( \alpha \)-particle are equal, we set their wavelengths equal: \[ \frac{h}{m_{{p}} v_{{p}}} = \frac{h}{m_{\alpha} v_{\alpha}}. \] Simplifying this gives: \[ \frac{v_{{p}}}{v_{\alpha}} = \frac{m_{\alpha}}{m_{{p}}}. \] 
Step 3: The mass of the \( \alpha \)-particle is approximately 4 times the mass of the proton (\( m_{\alpha} = 4m_{{p}} \)). Thus, the ratio of the velocities is: \[ \frac{v_{{p}}}{v_{\alpha}} = \frac{4}{1}. \] 
Step 4: Therefore, the ratio of their velocities is \( 4 : 1 \). \[ \boxed{4 : 1}. \]