Question

An electron and an alpha particle have the same de- broglie wavelength associated with them. How are their kinetic energies related to each other

Answer 1

K.E = 1/2$mv^{2}$

Now, divide and multiply R.H.S by "m"

K.E = 1/2m $P^{2}$ ...(1)

and, $λ_{e} = λ_{[tex]</p><p>from 1</p><p>\frac{K.E}{K.E}  = \frac{1}{2} \frac{P^{2}}{2_{me}}  \frac{\frac{1}{2} \frac{P^{2}}{2_{m∝}}}{}$}[/tex]

$\frac{K.E}{K.E}  = \frac{m_{∝} }{ m_{e}}$

{ m_{e}} < {m_{∝} }

So, $K_{e} > K_{alpha}$

Answer 2

An electron and alpha particle have the same kinetic energy. ... An electron and a proton have the same de Broglie wavelength associated with them. Write the relation between de-Broglie wavelength (λ) associated with the electron and its kinetic energy E.........