If the de Broglie wavelength of an electron is $\lambda_e$, then the exact wavelength of an alpha pa

Question

If the de Broglie wavelength of an electron is $\lambda_e$, then the exact wavelength of an alpha particle moving with one-fourth the velocity of this electron would be approximately (Assume the mass of a proton $\approx 1840 m_e$)

Professor PC Thomas Classes · Answer

(a) : By de Broglie's equation, $\lambda = \frac{h}{mv}$. Taking the ratio, we get $\frac{\lambda_e}{\lambda_\alpha} = \frac{m_\alpha v_\alpha}{m_e v_e}$. The mass of an alpha particle is approximately 4 times that of a proton (or a neutron), so $m_\alpha \approx 4 	imes 1840 m_e = 7360 m_e$. The given velocity is $v_\alpha = \frac{v_e}{4}$. Substituting these values: $\frac{\lambda_e}{\lambda_\alpha} = \frac{7360 m_e 	imes (v_e/4)}{m_e 	imes v_e} = \frac{1840 m_e v_e}{m_e v_e} = 1840$. Therefore, $\lambda_\alpha = \frac{\lambda_e}{1840}$.