Answer:
5.3 A
Explanation:
The orbital radius for the generic nth-level in the hydrogen atom is given by
[tex]a_n = n^2 a_0[/tex]
where:
[tex]a_0 = \frac{\epsilon_0 h^2}{\pi m_e e^2}[/tex]
is the Bohr radius, with
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] being the vacuum permittivity
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
[tex]m_e = 9.11\cdot 10^{-31} kg[/tex] is the electron mass
[tex]e=1.6\cdot 10^{-19} C[/tex] is the electron charge
Substituting all this numbers into the formula, we find
[tex]a_0 = 5.3\cdot 10^{-10} m = 5.3 A[/tex]
and since
n = 1
the radius of the hydrogen atom for the first principal quantum number is
[tex]a_1 = 1^2 a_0 = 1 \cdot (5.3 A)=5.3 A[/tex]
