Question

A singly ionized helium ion has only one electron and is denoted He⁺. What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

Short answer

The radius of ionised Helium is 0.5 times that of the Hydrogen atom.

Step-by-step solution

What is bohr’s theory explains?

The spectrum of hydrogen atoms is explained by an atomic structure theory. It is assumed that the electron orbiting the nucleus can only exist in particular energy states, with each transition resulting in the emission or absorption of a quantum of radiation.

Given information and Formula to be used

Given: Helium atom is singly ionized which means it now has a hydrogen like structure with just one electron.

Consider the formula for the Bohr’s radius as follows:

\({r_n} = \frac{{{n^2}{a_B}}}{Z}\)

Here, r_n is the Radiusnis the energy level, a_BBohr's radius, and Nis the number of proton.

Consider the value of \({a_B} = 0.59 \times {10^{ - 10}}\;{\rm{m}}\).

Determine the radius of ionized Helium atom with that of the Bohr radius of Hydrogen atom. 

Radius of hydrogen atom is calculated as:

For Hydrogen:

\(\begin{array}{c}{r_n} = \frac{{{n^2}{a_B}}}{Z}\\ = \frac{{{{\left( 1 \right)}^2}\left( {{a_B}} \right)}}{1}\\ = {a_B}\end{array}\)

Also,

\(\begin{array}{c}{r_n} = \frac{{1\left( {{a_B}} \right)}}{2}\\ = 0.5{a_B}\end{array}\)

Substitute the values and solve for the ionized helium as:

\(\begin{array}{c}{r_n} = \frac{{{n^2}{a_B}}}{Z}\\ = \frac{{{{\left( 1 \right)}^2}\left( {0.529 \times {{10}^{ - 10}}} \right)}}{2}\;{\rm{m}}\\ = 2.645 \times {10^{ - 11}}\;{\rm{m}}\end{array}\)

Hence, the radius of ionised Helium is 0.5 times that of the Hydrogen atom.