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

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Solution

Kinetic energy of particle of mass m having momentum p is,

K . E = \frac{1}{2} \frac{p^{2}}{m} \Rightarrow p = \sqrt{2 mK}

De-Broglie wavelength,

λ = \frac{h}{p} = \frac{h}{\sqrt{2 mK}}

∴

p = \frac{h}{λ}

....(i)
and,

K = \frac{h^{2}}{2 {mλ}^{2}}

....(ii)

If

λ

is constant, then from equation (i), p = constant.

i.e.,

m_{p} v_{p} = m_{e} v_{e}

\frac{v_{p}}{v_{e}} = \frac{m_{e}}{m_{p}} < 1 or v_{p} < v_{e}

λ

is constant, then from (ii),

K \propto \frac{1}{m}

∴

\frac{K_{p}}{K_{e}} = \frac{m_{e}}{m_{p}} < 1

K_{p} < K_{e}

It means the velocity of electron is greater than that of proton. Kinetic energy of electron is greater than that of proton.

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