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Chemistry
An electron has kinetic energy of 2.14 × 10-22J Its de-Broglie wavelength will be nearly (me = 9.1 × 10-31 kg)
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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
A
$9.28 \times 10^{-4} m$
11%
B
$9.28 \times 10^{-7} m$
21%
C
$9.28 \times 10^{-8} m$
57%
D
$9.28 \times 10^{-10} m$
11%
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$.