Radius of first orbit in H -atom is ' $\mathrm{a}_0$ ' Then, de-Broglie wavelength of electron in the third orbit is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Radius for } \mathrm{n}^{	ext {th }} 	ext { orbit, } \mathrm{r}_{\mathrm{n}}=\mathrm{a}_0 	imes \mathrm{n}^2 \ & 	ext { For third orbit, } \ & \mathrm{r}_3=\mathrm{a}_0 	imes 3^2 \ & \quad=9 \mathrm{a}_0 \ & 	ext { Also, } \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \ & \Rightarrow \quad \mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}}=\frac{3 \mathrm{~h}}{2 \pi 	imes 9 \mathrm{a}_0} \ & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{3 \mathrm{~h}} 	imes 2 \pi 	imes 9 \mathrm{a}_0=6 \pi \mathrm{a}_0\end{aligned}$ /

Radius of first orbit in H -atom is ' $\mathrm{a}_0$ ' Then, de-Broglie wavelength of electron in the third…

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