The de-Broglie wavelength of an electron (mas | vedclass

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(B) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Given values are:$h = 6.6 	imes 10^{-34} \ J \cdot s$$m = 1 \

Vedclass · Accepted Answer

$8.25 	imes 10^{-11} \ m$

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(B) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Given values are:$h = 6.6 	imes 10^{-34} \ J \cdot s$$m = 1 	imes 10^{-30} \ kg$$K = 200 \ eV = 200 	imes 1.6 	imes 10^{-19} \ J = 3.2 	imes 10^{-17} \ J$.Substituting these values into the formula:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes (1 	imes 10^{-30}) 	imes (3.2 	imes 10^{-17})}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{6.4 	imes 10^{-47}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{8 	imes 10^{-23.5}}$Using the calculation $\sqrt{64 	imes 10^{-48}} = 8 	imes 10^{-24}$, we get:$\lambda = \frac{6.6 	imes 10^{-34}}{8 	imes 10^{-24}} = 0.825 	imes 10^{-10} \ m = 8.25 	imes 10^{-11} \ m$.

The de-Broglie wavelength of an electron (mass $= 1 \times 10^{-30} \ kg$, charge $= 1.6 \times 10^{-19} \ C$) with a kinetic energy of $200 \ eV$ is (Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$):

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