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Q. What will be the kinetic energy of an electron having de-Broglie wavelength $ 2\overset{o}{\mathop{\text{A}}}\, $ ?

Rajasthan PETRajasthan PET 2011

Solution:

$ \lambda =\frac{12.27}{\sqrt{V}} $
$ V=\frac{{{(12.27)}^{2}}}{{{\lambda }^{2}}}=\frac{150}{{{\lambda }^{2}}} $
$ E=V\,eV=37.5\,eV $