For what kinetic energy of a proton, will the associated de-Broglie wavelength be 16.5 nm?

Solution

Given,
De-broglie wavelength, λ = 16.5 nm = 16.5 x 10^–9 m
Mass of proton m = 1.6 x 10^–27 kg

Using de-Broglie equation,

$λ = \frac{h}{mv}$

$\Rightarrow$ $v = \frac{h}{mλ}$
$∴$ $K . E . = \frac{1}{2} {mv}^{2} = \frac{1}{2} m . \frac{h^{2}}{m^{2} λ^{2}}$
$K . E . = \frac{h^{2}}{2 {mλ}^{2}}$
$\Rightarrow$ $= \frac{6.63 \times 10^{- 34} \times 6.63 \times 10^{- 34}}{2 \times 1.6 \times 10^{- 27} \times 16.5 \times 10^{- 9} \times 16.5 \times 10^{- 9}}$

$\Rightarrow$ $K . E . = \frac{6.63 \times 6.63 \times 10^{- 34 - 34 + 27 + 9 + 9}}{2 \times 1.6 \times 16.5 \times 16.5}$

i.e., $K . E . = 0.05045 \times 10^{- 68 + 45} = 5.045 \times 10^{- 2} \times 10^{- 23} K . E . = 5.045 \times 10^{- 25} J .$

is the calculated kinetic energy of proton.

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For what kinetic energy of a proton, will the associated de-Broglie wavelength be 16.5 nm?

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