Search any question & find its solution
Question:
Answered & Verified by Expert
The kinetic energy of an electron is 5 eV . Calculate the de-Broglie wavelength associated with it $\left(h=6.6 \times 10^{-34} \mathrm{Js}\right.$, $m_e=9.1 \times 10^{-31} \mathrm{~kg}$ )
Options:
Solution:
1620 Upvotes
Verified Answer
The correct answer is:
$5.47 Å$
Wavelength associated with an electron
$\lambda=\frac{h}{\sqrt{2 m E}}$
$\begin{aligned} & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{19}}} \\ & =5.47 \times 10^{-10} \mathrm{~m}=5.47 Å\end{aligned}$
$\lambda=\frac{h}{\sqrt{2 m E}}$
$\begin{aligned} & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{19}}} \\ & =5.47 \times 10^{-10} \mathrm{~m}=5.47 Å\end{aligned}$
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.