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Calculate the kinetic energy of the electron having wavelength $1 \mathrm{~nm}$.
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Verified Answer
The correct answer is:
$1.5 \mathrm{eV}$
de broglie wavelength,
$\lambda=\frac{h}{\sqrt{® m k}}$
or $k=\frac{h^{\circledast}}{๑ m \lambda^{\circledast}}$
$=\frac{\left(6.67 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(1 \times 10^{-9}\right)^2}=1.5 \mathrm{eV}$.
$\lambda=\frac{h}{\sqrt{® m k}}$
or $k=\frac{h^{\circledast}}{๑ m \lambda^{\circledast}}$
$=\frac{\left(6.67 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(1 \times 10^{-9}\right)^2}=1.5 \mathrm{eV}$.
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