Atomic Structure Result Question 12-1
12. The radius of the second Bohr orbit for hydrogen atom is (Planck’s constant $(h)=6.6262 \times 10^{-34} \mathrm{Js}$; mass of electron $=9.1091 \times 10^{-31} \mathrm{~kg} ;$ charge of electron $(e)=1.60210 \times 10^{-19} \mathrm{C}$; permitivity of vacuum $\left.\left(\epsilon_0\right)=8.854185 \times 10^{-12} \mathrm{~kg}^{-1} \mathrm{~m}^{-3} \mathrm{~A}^2\right)$
(2017 Main)
(a) $1.65 $ $Å$
(b) $4.76 $ $Å$
(c) $0.529 $ $Å$
(d) $2.12 $ $Å$
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Answer:
Correct Answer: 12. ( d )
Solution:
- Bohr radius $\left(r_n\right)=\epsilon_0 n^2 h^2$
$\begin{aligned} r_n & =\dfrac{n^2 h^2}{4 \pi^2 m e^2 k Z} \\ k & =\dfrac{1}{4 \pi \epsilon_0} \\ \therefore \quad r_n & =\dfrac{n^2 h^2 \epsilon_0}{\pi m e^2 Z}=n^2 \dfrac{a_0}{Z} \end{aligned}$
where, $m=$ mass of electron
$e=$ charge of electron
$h=$ Planck’s constant
$k=$ Coulomb constant
$r_n=\dfrac{n^2 \times 0.53}{Z} $ $Å$
Radius of $n^{\text {th }}$ Bohr orbit for $H$-atom $=0.53 n^2 $ $Å$ $\quad \quad [Z=1$ for $H$ -atom $]$
$\therefore \quad $ Radius of $2^{\text {nd }}$ Bohr orbit for $H$-atom $=0.53 \times(2)^2=2.12 $ $Å$
BETA
