If the radius of the first Bohr orbit is r,th | vedclass

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(C) According to Bohr's second postulate,the angular momentum is given by $\frac{nh}{2 \pi} = mvr_n$.From this,the de-Broglie wavelength $\lambda_n$ i

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$8 \pi r$

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(C) According to Bohr's second postulate,the angular momentum is given by $\frac{nh}{2 \pi} = mvr_n$.From this,the de-Broglie wavelength $\lambda_n$ is $\lambda_n = \frac{h}{mv} = \frac{2 \pi r_n}{n}$.We know that the radius of the $n^{	ext{th}}$ Bohr orbit is given by $r_n = r_1 	imes n^2$,where $r_1 = r$.For the $4^{	ext{th}}$ orbit $(n = 4)$,the radius is $r_4 = r 	imes 4^2 = 16r$.Substituting these values into the wavelength formula:$\lambda_4 = \frac{2 \pi r_4}{4} = \frac{2 \pi 	imes (16r)}{4} = 8 \pi r$.

If the radius of the first Bohr orbit is $r$,then the de-Broglie wavelength of the electron in the $4^{\text{th}}$ orbit will be:

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