Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1395 Upvotes
Verified Answer
The correct answer is:
1104.3
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 \(=\) ?
\(\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m(\mathrm{KE})}}\left(\text { here, } h=6.6 \times 10^{-34} \mathrm{Js}\right) \\
\lambda & =\frac{6.63 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \times 9.0 \times 10^{-31} \mathrm{~kg} \times 2 \times 10^{-25} \mathrm{~J}}} \\
\lambda & =1.1043 \times 10^{-6} \mathrm{~m}=\frac{1.1043 \times 10^{-6}}{10^{-9}} \mathrm{~nm} \\
& =1104.3 \mathrm{~nm}
\end{aligned}\)
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 \(=\) ?
\(\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m(\mathrm{KE})}}\left(\text { here, } h=6.6 \times 10^{-34} \mathrm{Js}\right) \\
\lambda & =\frac{6.63 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \times 9.0 \times 10^{-31} \mathrm{~kg} \times 2 \times 10^{-25} \mathrm{~J}}} \\
\lambda & =1.1043 \times 10^{-6} \mathrm{~m}=\frac{1.1043 \times 10^{-6}}{10^{-9}} \mathrm{~nm} \\
& =1104.3 \mathrm{~nm}
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.