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(D) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Rearranging for energy $E$,we get $E = \frac{h^2}

Vedclass · Accepted Answer

$8 	imes 10^{-25} \ 	ext{J}$

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(D) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Rearranging for energy $E$,we get $E = \frac{h^2}{2m\lambda^2}$.Given values: $h = 6.6 	imes 10^{-34} \ 	ext{Js}$,$m = 9.1 	imes 10^{-31} \ 	ext{kg}$,and $\lambda = 5500 \ 	ext{Å} = 5.5 	imes 10^{-7} \ 	ext{m}$.Substituting these values into the formula:$E = \frac{(6.6 	imes 10^{-34})^2}{2 	imes 9.1 	imes 10^{-31} 	imes (5.5 	imes 10^{-7})^2}$$E = \frac{43.56 	imes 10^{-68}}{18.2 	imes 10^{-31} 	imes 30.25 	imes 10^{-14}}$$E = \frac{43.56 	imes 10^{-68}}{550.55 	imes 10^{-45}}$$E \approx 0.0791 	imes 10^{-23} \ 	ext{J} \approx 7.91 	imes 10^{-25} \ 	ext{J}$.Rounding to the nearest provided option,we get $8 	imes 10^{-25} \ 	ext{J}$.

If an electron has an energy such that its de-Broglie wavelength is $5500 \ \text{Å}$,then the energy value of that electron is $(h = 6.6 \times 10^{-34} \ \text{Js}, m_e = 9.1 \times 10^{-31} \ \text{kg})$.

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