The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle

Question

The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2 E$, is $n: 1$. The value of $n$ is $\_\_\_\_$ .
(Assume mass of proton $=$ mass of neutron) :

Professor PC Thomas Classes · Answer

The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$, where $m$ is mass and $K$ is kinetic energy. Let the mass of a proton be $m_p$. A deuteron has a mass $m_d \approx 2m_p$ and kinetic energy $K_d = E$. Its wavelength is $\lambda_d = \frac{h}{\sqrt{2(2m_p)E}} = \frac{h}{\sqrt{4m_p E}}$. An alpha particle has a mass $m_\alpha \approx 4m_p$ and kinetic energy $K_\alpha = 2E$. Its wavelength is $\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2E)}} = \frac{h}{\sqrt{16m_p E}}$. The ratio is $\frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{16}}{\sqrt{4}} = \frac{4}{2} = 2$. Thus, $n = 2$.