Question

What is the Bohr radius of a hydrogen atom?

Hint:

The Bohr radius is a fundamental constant and represents the distance between the electron and the nucleus in the hydrogen atom’s ground state.

Answer 1

Step 1: Understanding the Bohr radius.
The Bohr radius \( a_0 \) is the radius of the smallest orbit around the nucleus in a hydrogen atom. It is given by the formula: \[ a_0 = \frac{4 \pi \epsilon_0 h^2}{m_e e^2} \] where: - \( \epsilon_0 \) is the permittivity of free space, - \( h \) is Planck’s constant, - \( m_e \) is the electron mass, - \( e \) is the electron charge. Step 2: Numerical value.
The numerical value for the Bohr radius is approximately \( 0.53 \, \text{Å} \).
Step 3: Conclusion.
Thus, the Bohr radius of a hydrogen atom is \( a_0 \approx 0.53 \, \text{Å} \).

Answer 2

Step 1: The Bohr radius is the size of the smallest possible orbit an electron can have around a hydrogen nucleus.

Step 2: It comes from the formula a₀ = 4πε₀h² ÷ (mₑe²), combining constants like Planck's constant, electron mass, and electron charge.

Step 3: Working this out gives a fixed value.

Step 4: So the Bohr radius of hydrogen is about 0.53 Å.

Answer 3

Derivation from Bohr's postulates.
Bohr assumed the electron's centripetal force is supplied by Coulomb attraction: \( \dfrac{m_e v^2}{r} = \dfrac{e^2}{4\pi\epsilon_0 r^2} \), and that angular momentum is quantised as \( m_e v r = \dfrac{nh}{2\pi} \).
Eliminating \( v \) between these two relations and solving for \( r \) at \( n = 1 \) gives \[ a_0 = \frac{4\pi\epsilon_0 h^2}{m_e e^2} \] Substituting the known values of \( \epsilon_0, h, m_e, e \) evaluates this to \( a_0 \approx 0.53 \, \text{Å} \), the Bohr radius.