If a_0 is the Bohr radius,then the de-Broglie | vedclass

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(A) According to Bohr's postulate,the angular momentum is given by $mvr = \frac{nh}{2\pi}$ --- $(1)$From de-Broglie's relation,the momentum is $p = mv

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$6\pi a_0$

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(A) According to Bohr's postulate,the angular momentum is given by $mvr = \frac{nh}{2\pi}$ --- $(1)$From de-Broglie's relation,the momentum is $p = mv = \frac{h}{\lambda}$ --- $(2)$Substituting the value of $mv$ from equation $(2)$ into equation $(1)$:$\frac{h}{\lambda} 	imes r_n = \frac{nh}{2\pi}$$\lambda = \frac{2\pi r_n}{n}$For the second excited state,the principal quantum number $n = 3$.The radius of the $n^{th}$ orbit is $r_n = n^2 a_0$. For $n = 3$,$r_3 = 3^2 a_0 = 9 a_0$.Substituting these values into the expression for $\lambda$:$\lambda = \frac{2\pi 	imes 9 a_0}{3} = 6\pi a_0$.

If $a_0$ is the Bohr radius,then the de-Broglie wavelength of an electron revolving in the second excited state of the $H$ atom will be:

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