What is the energy of a neutron with a de Bro | vedclass

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h = 6.63 	imes 10^{-34} \ 	ext{J} \cdot 	ext{s}$.Given $\lambda = 1 \ \mat

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$8.13 	imes 10^{-2} \ 	ext{eV}$

Vedclass · Answer

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h = 6.63 	imes 10^{-34} \ 	ext{J} \cdot 	ext{s}$.Given $\lambda = 1 \ \mathring{A} = 10^{-10} \ 	ext{m}$.The momentum $p = \frac{h}{\lambda} = \frac{6.63 	imes 10^{-34}}{10^{-10}} = 6.63 	imes 10^{-24} \ 	ext{kg} \cdot 	ext{m/s}$.The kinetic energy $E$ of a neutron is given by $E = \frac{p^2}{2m_n}$,where $m_n \approx 1.675 	imes 10^{-27} \ 	ext{kg}$.$E = \frac{(6.63 	imes 10^{-24})^2}{2 	imes 1.675 	imes 10^{-27}} \ 	ext{J} \approx \frac{43.96 	imes 10^{-48}}{3.35 	imes 10^{-27}} \ 	ext{J} \approx 1.312 	imes 10^{-20} \ 	ext{J}$.To convert this to electron volts $(	ext{eV})$,divide by $1.6 	imes 10^{-19} \ 	ext{J/eV}$:$E = \frac{1.312 	imes 10^{-20}}{1.6 	imes 10^{-19}} \ 	ext{eV} \approx 0.082 \ 	ext{eV} = 8.2 	imes 10^{-2} \ 	ext{eV}$.Rounding to the nearest provided option,the correct answer is $8.13 	imes 10^{-2} \ 	ext{eV}$.

What is the energy of a neutron with a de Broglie wavelength of $1 \ \mathring{A}$?

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