Question

The radius of the second Bohr orbit for hydrogen atom in angstrom is:
$\text{(Planck's constant,}h=6.262\times {10}^{-34}\text{Js}:$ $\text{Mass of electron}=9.1091\times {10}^{-31}\text{kg};$ $\text{Charge of electron,}e=1.60210\times {10}^{-19}\text{C};$ $\text{permittivity of vacuum,}{\epsilon }_0=8.854185\times {10}^{-12}{\text{kg}}^{-1}{\text{m}}^{-3}{\text{A}}^2)$

• A. $1.65$
• B. $4.76$
• C. $0.529$
• D. $2.12$

Answer 1

The correct option is D $2.12$

$\text{Radius of}{n}^{th}\text{Bohr orbit in}H\text{atom}=0.53\times \frac{n^2}{Z}$
$\text{For hydrogen}Z=1$
$\text{Radius of}{2}^{nd}\text{Bohr orbit in}H\text{atom}=0.53\times \frac{2^2}{1}=2.12$
$\text{Hence option (d) is the answer.}$