Atomic Structure Question 171

Question: An electron has kinetic energy $ 2.8\times {10^{-23}}J $ . de-Broglie wavelength will be nearly $ (m_{e}=9.1\times {10^{-31}}kg) $

Options:

A) $ 9.28\times {10^{-4}},m $

B) $ 9.28\times {10^{-7}},m $

C) $ 9.28\times {10^{-8}},m $

D) $ 9.28\times {10^{-10}},m $

Show Answer

Answer:

Correct Answer: C

Solution:

Formula for de-Broglie wavelength is $ \lambda =\frac{h}{p} $ or $ \lambda =\frac{h}{mv}\Rightarrow eV=\frac{1}{2}mv^{2} $ or $ \nu =\sqrt{\frac{2eV}{m}} $

$ \lambda =\frac{h}{\sqrt{2meV}} $

$ =\frac{6.62\times {10^{-34}}}{\sqrt{2\times 9.1\times {10^{-31}}\times 2.8\times {10^{-23}}}} $

$ \lambda =9.28\times {10^{-8}}meter $ .



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