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Calculate the de-Broglie's wavelength of an electron residing in the 2nd Bohr's orbit of a hydrogen atom. (Bohr's radius, $a_0=0.529 Å$ )
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Verified Answer
The correct answer is:
$2.116 \pi Å$
Bohr’s radius is equal to most probable distance between the nuclear and electron in H-atom in its ground states.
According to de-Broglie wavelength, the allowed any stationary orbit
i.e. $n \lambda=2 \pi r$
$\begin{aligned} & n \lambda=2 \pi\left(5.29 \times 10^{-11}\right)\left(\frac{n^2}{Z}\right) \\ & \lambda=2 \pi\left(5.29 \times 10^{-11}\right) \times \frac{n}{Z}...(i)\end{aligned}$
where, $r=5.29 \times 10^{-11}\left(\frac{n^2}{Z}\right)$
Putting $Z=1$
(For H-atom)
$n=2$
(For 2-orbit)
in Eq. (i), we get
$$
\begin{aligned}
& \lambda=4 \pi\left(5.29 \times 10^{-11}\right) \mathrm{m} \\
& \lambda=4 \times 0.529 \pi \times 10^{-10} \mathrm{~m} \\
& \lambda=2.116 \pi Å
\end{aligned}
$$
Hence, de-Broglie wavelength in 2nd orbit is $2.116 \pi Å$.
According to de-Broglie wavelength, the allowed any stationary orbit
i.e. $n \lambda=2 \pi r$
$\begin{aligned} & n \lambda=2 \pi\left(5.29 \times 10^{-11}\right)\left(\frac{n^2}{Z}\right) \\ & \lambda=2 \pi\left(5.29 \times 10^{-11}\right) \times \frac{n}{Z}...(i)\end{aligned}$
where, $r=5.29 \times 10^{-11}\left(\frac{n^2}{Z}\right)$
Putting $Z=1$
(For H-atom)
$n=2$
(For 2-orbit)
in Eq. (i), we get
$$
\begin{aligned}
& \lambda=4 \pi\left(5.29 \times 10^{-11}\right) \mathrm{m} \\
& \lambda=4 \times 0.529 \pi \times 10^{-10} \mathrm{~m} \\
& \lambda=2.116 \pi Å
\end{aligned}
$$
Hence, de-Broglie wavelength in 2nd orbit is $2.116 \pi Å$.
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