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# The angular momentum of an electron in Bohr’s orbit of He+ is 3.1652*10^-34 kgm^2/sec. what is wave number in terms of rydbergs constant (R) of the spectral line emitted if an electron falls from this level to first excited state ( using h=6.626*10^-34 Joule second. )

Eshan
3 years ago
Dear student,

Angular momentum in Bohr’s orbit is quantized and is equal to

$\dpi{80} \dfrac{nh}{2\pi}=3.1652\times 10^{-34}$
$\dpi{80} \implies n=3$

The electrons falls from this level to first excited state, that is, n=2.
Hence the wave number of emitted line=$\dpi{80} R(\dfrac{1}{2^2}-\dfrac{1}{3^2})=\dfrac{5R}{36}$