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(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the mo

Vedclass · Accepted Answer

$6.625 	imes 10^{-24} \ kg \ m/s$

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(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Rearranging for momentum: $p = \frac{h}{\lambda}$.Given: $\lambda = 1 \ \mathring{A} = 10^{-10} \ m$ and $h = 6.625 	imes 10^{-34} \ J \ s$.Substituting the values: $p = \frac{6.625 	imes 10^{-34} \ J \ s}{10^{-10} \ m} = 6.625 	imes 10^{-24} \ kg \ m/s$.

Calculate the momentum of a particle which has a wavelength of $1 \ \mathring{A}$.

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