Physics - Modern Physics Question with Solution | TestHub

The formula to calculate the de Broglie wavelength of a particle is given by

$\lambda =\frac{h}{\sqrt{2mK}} ...\left(1\right)$

Now, the masses of the given particles are interrelated by the following relations:

$\begin{array}{rcl}{m}_e& \approx & \frac{m_p}{1837}\\ m_{\alpha }& =& 4m_p\end{array}$

The de Broglie wavelength for electron is given by

$\begin{array}{rcl}{\lambda }_e& =& \frac{h}{\sqrt{2m_e\left(4K\right)}}\\ & =& \frac{h}{\sqrt{{\displaystyle \frac{8}{1837}}{m}_{p}K}} ...\left(2\right)\end{array}$

The de Broglie wavelength for proton is given by

$\begin{array}{rcl}{\lambda }_p& =& \frac{h}{\sqrt{2m_p\left(2K\right)}}\\ & =& \frac{h}{\sqrt{4m_{p}K}} ...\left(3\right)\end{array}$

And, the de Broglie wavelength of alpha particle is given by

$\begin{array}{rcl}{\lambda }_{\alpha }& =& \frac{h}{\sqrt{2m_{\alpha }K}}\\ & =& \frac{h}{\sqrt{2\left(4m_p\right)K}}\\ & =& \frac{h}{\sqrt{8m_{p}K}} ...\left(4\right)\end{array}$

From equation (2), (3) and (4), it can be concluded that ${\lambda }_e>{\lambda }_p>{\lambda }_{\alpha }$.