Physics - Modern Physics Question with Solution | TestHub

de brogile wavelength, $\lambda =\frac{h}{p}$ or $\lambda =\frac{h}{mv}$ $\therefore \frac{{\lambda }_p}{{\lambda }_e}=\frac{m_e{v}_e}{m_p{v}_p}$ $\Rightarrow m_p=\frac{m_e{v}_e}{v_p}\times \frac{{\lambda }_e}{{\lambda }_p}$ Here, $m_e=9.1\times 10^{-31}\mathrm{kg},v_p=3v_e$ and $\frac{{\lambda }_p}{{\lambda }_e}=1.814\times 10^{-4}$ $\therefore m_p=\frac{9.1\times 10^{-31}}{1.814\times 10^{-4}\times 3}=1.672\times 10^{-27}\mathrm{kg}$ Thus, the particle is neutron.