According to de-Broglie,the de-Broglie wavele | vedclass

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(C) According to the de-Broglie hypothesis for Bohr's orbits,the circumference of the orbit is an integral multiple of the de-Broglie wavelength:$2 \p

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) According to the de-Broglie hypothesis for Bohr's orbits,the circumference of the orbit is an integral multiple of the de-Broglie wavelength:$2 \pi r = n \lambda$Where $n$ is the principal quantum number,$r$ is the radius of the orbit,and $\lambda$ is the de-Broglie wavelength.Given: $r = 5.3 	imes 10^{-11} \ m$ and $\lambda = 10^{-10} \ m$.Substituting the values:$n = \frac{2 \pi r}{\lambda} = \frac{2 	imes 3.14 	imes 5.3 	imes 10^{-11}}{10^{-10}}$$n = \frac{33.284 	imes 10^{-11}}{10^{-10}} = 3.3284 \approx 3$Thus,the principal quantum number is $3$.

According to de-Broglie,the de-Broglie wavelength for an electron in an orbit of radius $5.3 \times 10^{-11} \ m$ of a hydrogen atom is $10^{-10} \ m$. The principal quantum number for this electron is:

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