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  4. According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is 10-9 m . The principle quantum number for this electron is

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Q. According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is $10^{-9} m .$ The principle quantum number for this electron is

Dual Nature of Radiation and Matter

Solution:

$2 \pi r n=\lambda \Rightarrow n=\frac{\lambda}{2 \pi r}$
$=\frac{10^{-9}}{2 \times 3.14 \times 5.13 \times 10^{-11}}=3$