9 - N u m e r i c a l 7

Problem

Calculate the radius of the innermost orbital level ( $n=1$ ) of the hydrogen atom.

Energy level ( $n$ ): $1$
Planck’s constant ( $h$ ): $6.626 \times 10^{-34} \, \mathrm{Js}$
Electron mass ( $m$ ): $9.109 \times 10^{-31} \, \mathrm{kg}$
Coulomb’s constant ( $k$ ): $8.988 \times 10^{9} \, \mathrm{Nm}^{2}/\mathrm{C}^{2}$
Charge on electron ( $e$ ): $1.602 \times 10^{-19} \, \mathrm{C}$
$\pi$ : $3.14$

To Find:
Radius of the innermost orbital ( $r_1$ ).

Bohr’s Model for Orbital Radius:
The radius of the $n$ -th orbital in the hydrogen atom is given by:

r_n = \frac{n^2 h^2}{4 \pi^2 m k e^2}

Where:

Innermost Orbital Radius ( $n=1$ ):
Substituting $n=1$ simplifies the formula to:

r_1 = \frac{h^2}{4 \pi^2 m k e^2}

r_1 = \frac{h^2}{4 \pi^2 m k e^2}

r_1 = \frac{(6.626 \times 10^{-34})^2}{4 (3.14)^2 (9.109 \times 10^{-31}) (8.988 \times 10^{9}) (1.602 \times 10^{-19})^2}

h^2 = (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67}

4 \pi^2 = 4 (3.14)^2 = 39.4784

m = 9.109 \times 10^{-31}

k = 8.988 \times 10^{9}

e^2 = (1.602 \times 10^{-19})^2 = 2.566 \times 10^{-38}

\text{Denominator: } 39.4784 \cdot 9.109 \times 10^{-31} \cdot 8.988 \times 10^{9} \cdot 2.566 \times 10^{-38} = 8.27 \times 10^{-48}

r_1 = \frac{4.39 \times 10^{-67}}{8.27 \times 10^{-48}}

r_1 = 0.53 \times 10^{-10} \, \mathrm{m}

Convert to picometers ( $\mathrm{pm}$ ):

r_1 = 0.53 \, \mathrm{Å} = 53 \, \mathrm{pm}

The radius of the innermost orbital level of the hydrogen atom is:

r_1 = 0.53 \, \mathrm{Å} \, \text{or} \, 53 \, \mathrm{pm}