Question

Find ratio \( \left(\frac{\lambda_a}{\lambda_p}\right) \) of the de Broglie wavelength \( \lambda_a \) and \( \lambda_p \) associated respectively with an alpha particle and a proton,
(i) if they are moving with the same kinetic energy.
(ii) just after they are accelerated through the same potential difference.

Hint:

Always include both mass and charge when potential difference is given.

Answer 1

Concept: De Broglie wavelength: \[ \lambda = \frac{h}{p} \quad,\quad p = \sqrt{2mK} \] So: \[ \lambda = \frac{h}{\sqrt{2mK}} \]

Step 1: Alpha particle and proton masses \[ m_\alpha = 4m_p \] Case (i): Same kinetic energy \[ \lambda \propto \frac{1}{\sqrt{m}} \] So: \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} = \sqrt{\frac{m_p}{4m_p}} = \frac{1}{2} \] \[ \boxed{\left(\frac{\lambda_\alpha}{\lambda_p}\right)_1 = \frac{1}{2}} \] Case (ii): Same potential difference Kinetic energy: \[ K = qV \] Alpha particle charge = $2e$, proton charge = $e$ So: \[ K_\alpha = 2eV,\quad K_p = eV \] Now: \[ \lambda = \frac{h}{\sqrt{2mK}} \] \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p K_p}{m_\alpha K_\alpha}} \] Substitute: \[ = \sqrt{\frac{m_p (eV)}{(4m_p)(2eV)}} \] \[ = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \] \[ \boxed{\left(\frac{\lambda_\alpha}{\lambda_p}\right)_2 = \frac{1}{2\sqrt{2}}} \] Final Answer: \[ \text{(i) } \frac{1}{2}, \quad \text{(ii) } \frac{1}{2\sqrt{2}} \]