Question

Find the radius of a hydrogen atom in the n = 2 state according to Bohr’s theory.

Short answer

The radius of a hydrogen atom in the 2^nd state is \[2.116 \times {10^{ - 10}}\;{\rm{m }}{\rm{.}}\]

Step-by-step solution

Determine the Bohr’s theory

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

Consider the formula for the radius of the atomic orbit is as follows:

\[{r_B} = \frac{{{n^2}{a_B}}}{z}\]

Here, r_B is the radius of orbit, nis the energy level, a_B is the Bohr's radius and z is the number of proton.

Determine the radius of a hydrogen atom in the 2 state. 

Radius of hydrogen atom is calculated as

\[{r_B} = \frac{{{n^2}{a_B}}}{z}\]

Substitute the value in the above expression

\[\begin{array}{c}{r_B} = \frac{{{{\left( 2 \right)}^2}\left( {0.529 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}}{1}\\{r_B} = {\left( 2 \right)^2}\left( {0.529 \times {{10}^{ - 10}}\;{\rm{m}}} \right)\\{r_B} = 4\left( {0.529 \times {{10}^{ - 10}}\;{\rm{m}}} \right)\\{r_B} = 2.116 \times {10^{ - 10}}\;{\rm{m}}\end{array}\]

Hence, the radius of a hydrogen atom in 2 state is \[2.116 \times {10^{ - 10}}\;{\rm{m}}\].