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

We need to find the ratio of de Broglie wavelengths of deuteron (KE = E) and alpha particle (KE = 2E).

Formula: $$\lambda = \frac{h}{\sqrt{2mK}}$$

Deuteron: mass = 2m_p (1 proton + 1 neutron), KE = E

Alpha: mass = 4m_p (2 protons + 2 neutrons), KE = 2E

$$\frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{2m_\alpha \times 2E}}{\sqrt{2m_d \times E}} = \sqrt{\frac{4m_p \times 2E}{2m_p \times E}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$$

So n:1 = 2:1, meaning n = 2.