Physics - Modern Physics Question with Solution | TestHub

de Broglie wavelength $\lambda =\frac{h}{mv}$ or $\lambda \propto \frac{1}{m}$ (when $v$ is constant) As $m_{\alpha }>m_n>m_p>m_{\beta }\therefore {\lambda }_{\alpha }<{\lambda }_n<{\lambda }_p<{\lambda }_{\beta }$ Therefore, $\beta$-particle has the longest de Broglie wavelength.