A particle of mass 1 times 10^{-30} ,kg and e | vedclass

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(D) The de-Broglie wavelength is given by $\lambda = h / p$. Momentum of the particle is $p = h / \lambda = (6.6 	imes 10^{-34}) / (660 	imes 10^{-9

Vedclass · Accepted Answer

$3.1 	imes 10^{-6} \,eV$

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(D) The de-Broglie wavelength is given by $\lambda = h / p$. Momentum of the particle is $p = h / \lambda = (6.6 	imes 10^{-34}) / (660 	imes 10^{-9}) = 1 	imes 10^{-27} \,kg \cdot m/s$. Kinetic energy $K$ in terms of momentum $p$ and mass $m$ is $K = p^2 / (2m)$. Substituting the values: $K = (1 	imes 10^{-27})^2 / (2 	imes 1 	imes 10^{-30}) = (1 	imes 10^{-54}) / (2 	imes 10^{-30}) = 0.5 	imes 10^{-24} \,J$. To convert this into electron-volts $(eV)$,divide by $1.6 	imes 10^{-19} \,J/eV$: $K = (0.5 	imes 10^{-24}) / (1.6 	imes 10^{-19}) \,eV = 0.3125 	imes 10^{-5} \,eV = 3.125 	imes 10^{-6} \,eV$. Rounding to two significant figures,we get $K \approx 3.1 	imes 10^{-6} \,eV$.

$A$ particle of mass $1 \times 10^{-30} \,kg$ and electric charge $1.6 \times 10^{-19} \,C$ has a de-Broglie wavelength of $660 \,nm$. The kinetic energy of this particle is (Planck's constant,$h = 6.6 \times 10^{-34} \,J \cdot s$)

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