An electron is moving with a precise kinetic energy of $4.55 imes 10^{-25} \mathrm{~J}$. The accur

Question

An electron is moving with a precise kinetic energy of $4.55 	imes 10^{-25} \mathrm{~J}$. The accurate de Broglie wavelength associated with this electron is ($h = 6.626 	imes 10^{-34} \mathrm{~Js}, m_e = 9.1 	imes 10^{-31} \mathrm{~kg}$)

Professor PC Thomas Classes · Answer

(a) : The de Broglie wavelength expressed via kinetic energy is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$. Substituting the standard numerical values: $\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 4.55 	imes 10^{-25}}} = \frac{6.626 	imes 10^{-34}}{\sqrt{82.81 	imes 10^{-56}}}$. Since $2 	imes 4.55 = 9.1$, the denominator's square root simplifies cleanly to $\sqrt{9.1^2 	imes 10^{-56}} = 9.1 	imes 10^{-28}$. Thus, $\lambda = \frac{6.626 	imes 10^{-34}}{9.1 	imes 10^{-28}} \approx 0.728 	imes 10^{-6} \mathrm{~m} = 728 \mathrm{~nm} = 7280 \mathrm{~\AA}$.