Dual Nature Of Light Question 420
Question: The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometer is
Options:
A) 0.168 eV
B) 16.8 eV
C) 1.68 eV
D) 2.5 eV
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \lambda =\frac{h}{\sqrt{2mE}}\Rightarrow E=\frac{h^{2}}{2m{{\lambda }^{2}}} $ $ =\frac{{{(6.6\times {{10}^{-34}})}^{2}}}{2\times 9.1\times {{10}^{-31}}\times {{(0.3\times {{10}^{-9}})}^{2}}}=2.65\times {{10}^{-18}}J $ $ =16.8,eV $
BETA
