Question

Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is\(\frac{ 9}{x} x 10^{-10}\)m. The value of the x is _____.
(1eV = 1.6 x 10^-19J, \(\frac{1}{4π∈_0}\) = 9 x 10⁹ NM² / C² and electronic charge = 1.6 x 10^-19JC)

Answer 1

Correct Answer: 16

To find the value of \(x\) for the orbital radius of the electron, we start by understanding the energy needed to ionize a hydrogen atom. This energy (\(E\)) is given as 12.8 eV, which can be converted to joules: \(12.8 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 2.048 \times 10^{-18} \text{ J}\).

The energy of an electron in a hydrogen atom is given by the formula \(E = -\frac{k e^2}{2r}\), where \(k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{/C}^2\), \(e = 1.6 \times 10^{-19} \text{ C}\) and \(r\) is the orbital radius. To find \(r\), we equate this energy to the required ionization energy (magnitude):

\[\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2r} = 2.048 \times 10^{-18}\]

Simplify the left side:

\[\frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{2r} = 2.048 \times 10^{-18}\]

\[\frac{23.04 \times 10^{-29}}{r} = 2.048 \times 10^{-18}\]

Solving for \(r\):

\[r = \frac{23.04 \times 10^{-29}}{2.048 \times 10^{-18}}\]

\[r = 11.25 \times 10^{-11} \text{ m}\]

The given expression for \(r\) is \(\frac{9}{x} \times 10^{-10} \text{ m}\). Setting it equal to \(11.25 \times 10^{-11} \text{ m}\) gives:

\[\frac{9}{x} \times 10^{-10} = 11.25 \times 10^{-11}\]

\[\frac{9}{x} = 1.125\]

Solving for \(x\):

\[x = \frac{9}{1.125} = 8\]

The value of \(x\) is 8, which is different from the provided range (16,16). The computed value falls outside any range ambiguities, indicating either a calculation misfile or range misalignment in problem constraints.