Calculate the de-Broglie's wavelength associated with an electron of energy 200eV What will be the change in its wavelength - Physics - Current Electricity

Question

Calculate the de-Broglie's wavelength associated with an
electron of energy 200eV What will be the change in its wavelength
if the accelerating potential is increased to four times its
earlier value?

Vijay Yadav · Accepted Answer

Hello,

de - Broghe's wavelength is given by

$λ = \frac{h}{P}$

Where h = Plank constant

P = momentum

P = mv

P² = (mv)² = $\frac{2 \times m^{2} v^{2}}{2}$

$\Rightarrow P^{2} = 2 m (\frac{1}{2} {mv}^{2}) = 2 m * KE \Rightarrow P= \sqrt{2m (KE)}$

Given that KE = 200 eV

me = 9.11 × 10^–31 kg

e = 1.6 × 10^–19 C

$P = \sqrt{2 \times 9.11 \times 10^{- 31} \times 200 \times 1.6 \times 10^{- 19}}$

P = 7.63 × 10^–24 kg-m/s.

Hence, wavelength of the electron

$λ = \frac{h}{P} = \frac{6.63 \times 10^{- 34}}{7.63 \times 10^{- 24}} = 0.868 \times 10^{- 10} m λ = 8.68 \times 10^{- 9} m$

It the accelerating voltage of electrons is increased to four times then its K.E. also increases to four times.

Since $P α \sqrt{K.E.}$

Hence momentum is increased twice.

P' = 2P

So new wavelength of the electron

$λ' = \frac{h}{P'} = \frac{h}{2 P} = \frac{λ}{2} λ' = 4.34 \times 10^{- 9} m$

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Calculate the de-Broglie's wavelength associated with an electron of energy 200eV .What will be the change in its wavelength if the accelerating potential is increased to four times its earlier value?