Question

A proton and an electron have same d-Broglie wavelength which of them moves fast and which possess more K.E. Justify your answer.

Answer 1

Kinetic energy K of a particle of mass m having
Momentum p is , `K=1/2 (p^2)/m or p=sqrt(2mK)`
De-Broglie wavelength, `lambda=h/p=h/(sqrt(2mK))`
`:. P=h/lambda` ....(i) and `K=(h^2)/(2mlambda^2)`.....(ii)
If `lambda` is constant, then from (i), p=a constant
i.e., `m_pv_p=m_ev_e` or `(v_p)/(v_e)=(m_e)/(m_p) lt 1`
or `v_p lt v_e`
If `lambda` is constant, the form (ii), `Kprop1//m`
`(K_p)/(K_e)=(m_e)/(m_p) lt 1` or `K_p lt K_e`
It means the velocity of electron is greater than that of proton. Kinetic energy of electron is greater than that of proton.