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Q. An electron has kinetic energy of $2.14 \times 10^{-22}J$ Its de-Broglie wavelength will be nearly $(m_e = 9.1 \times 10^{-31}$ kg)

Structure of Atom

Solution:

$E = \frac{hc}{\lambda}$
or $\lambda = \frac{hc}{E}$
$ = \frac{6.62\times 10^{-34} \times 3 \times 10^8}{2.14 \times 10^{-22}}$
$ = 9.28\times 10^{-8}\,m$.