The radius of first Bohr orbit in hydrogen atom is

A
52.9 cm
B
52.9 nm
C
52.9 pm
D
$52.9 Å$

Solution:

$r = n^{2} [\frac{h^{2}}{4 π^{2} m (e^{2} / 4 π ε_{0})}] = n^{2} (\frac{ε_{0} h^{2}}{π m e^{2}}) = 1^{2} [\frac{(8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}) {(6.626 \times 10^{- 34} Js)}^{2}}{(3.14) (9.1 \times 10^{- 31} kg) {(1.602 \times 10^{- 19} C)}^{2}}]$

$= 5.29 \times 10^{- 11} m = 52.9 pm$

Solution:

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