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If the radius of first orbit of $\mathrm{H}$ atom is $a_0$, the deBroglie wavelength of an electron in the third orbit is
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Verified Answer
The correct answer is:
$6 \pi a_0$
$6 \pi a_0$
$r_n=a_0 n^2$
$$
\begin{aligned}
& r=a_0 \times(3)^2=9 a_0 \\
& m v r=\frac{n h}{2 \pi} \\
& m v=\frac{n h}{2 \pi \mathrm{r}}=\frac{3 h}{2 \pi \times 9 a_0}=\frac{h}{6 \pi a_0} \\
& \lambda=\frac{h}{m v}=\frac{h}{h} \times 6 \pi a_0=6 \pi a_0
\end{aligned}
$$
$$
\begin{aligned}
& r=a_0 \times(3)^2=9 a_0 \\
& m v r=\frac{n h}{2 \pi} \\
& m v=\frac{n h}{2 \pi \mathrm{r}}=\frac{3 h}{2 \pi \times 9 a_0}=\frac{h}{6 \pi a_0} \\
& \lambda=\frac{h}{m v}=\frac{h}{h} \times 6 \pi a_0=6 \pi a_0
\end{aligned}
$$
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