Calculate the
(a) momentum, and
(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

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Solution

Given,
Potential difference, V = 56 V

Energy of electron accelerated,

= 56 eV = 56 \times 1.6 \times 10^{- 19} J

(a) As, Energy,

E = \frac{p^{2}}{2 m}

[p = mv, E = \frac{1}{2} {mv}^{2}]

∴

p^{2} = 2 mE

\Rightarrow

p = \sqrt{2 mE}

\Rightarrow

p = \sqrt{2 \times 9 \times 10^{- 31} \times 56 \times 1.6 \times 10^{- 19}}

p = 4.02 \times 10^{- 24} kg {ms}^{- 1}

is the momentum of the electron.

(b) Now, using De-broglie formula we have,

p = \frac{h}{λ}

∴

λ = \frac{h}{p} = \frac{6.62 \times 10^{- 34}}{4.02 \times 10^{- 24}}

= 1.64 \times 10^{- 10} m = 0.164 \times 10^{- 9} m

i.e.,

λ = 0.164 nm,

is the De-broglie wavelength of the electron.

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