If the kinetic energy of an electron of mass \(9.0 \times 10^{-31} \mathrm{~kg}\) is \(2.0 \times 10^{-25}…
If the kinetic energy of an electron of mass \(9.0 \times 10^{-31} \mathrm{~kg}\) is \(2.0 \times 10^{-25} \mathrm{~J}\), the wavelength of the electron in \(\mathrm{nm}\) is approximately
1004.3
1204.3
1104.3
994.3
Solution
Given,
Kinetic energy of an electron \(=2.0 \times 10^{-25} \mathrm{~J}\)
Mass of electron \(=9.0 \times 10^{-31} \mathrm{~kg}\)
Wavelength of the electron \(=\) ?
\(\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m(\mathrm{KE})}}\left(\text { here, } h=6.6 \times 10^{-34} \mathrm{Js}\right) \\
\lambda & =\frac{6.63 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \times 9.0 \times 10^{-31} \mathrm{~kg} \times 2 \times 10^{-25} \mathrm{~J}}} \\
\lambda & =1.1043 \times 10^{-6} \mathrm{~m}=\frac{1.1043 \times 10^{-6}}{10^{-9}} \mathrm{~nm} \\
& =1104.3 \mathrm{~nm}
\end{aligned}\)