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Question

(C) According to de Broglie's hypothesis,the circumference of the $n^{th}$ orbit is equal to an integral multiple of the de Broglie wavelength: $2\pi

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) According to de Broglie's hypothesis,the circumference of the $n^{th}$ orbit is equal to an integral multiple of the de Broglie wavelength: $2\pi r_n = n\lambda$.Given: Wavelength $\lambda = 10^{-9} \ m$.The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = n^2 a_0$,where $a_0 = 0.529 	imes 10^{-10} \ m$.For $n=3$,$r_3 = 3^2 	imes 0.529 	imes 10^{-10} \ m = 9 	imes 0.529 	imes 10^{-10} \ m \approx 4.76 	imes 10^{-10} \ m$.Using the condition $n = \frac{2\pi r_n}{\lambda}$:$n = \frac{2 	imes 3.14159 	imes 4.76 	imes 10^{-10}}{10^{-9}} \approx 3$.Thus,the principal quantum number is $3$.

The wavelength for an electron in an orbit of a hydrogen atom is $10^{-9} \ m$. The principal quantum number for this electron is:

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