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Expert-verifiedFind the radius of a hydrogen atom in the n = 2 state according to Bohr’s theory.
The radius of a hydrogen atom in the 2nd state is \[2.116 \times {10^{ - 10}}\;{\rm{m }}{\rm{.}}\]
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.
Consider the formula for the radius of the atomic orbit is as follows:
\[{r_B} = \frac{{{n^2}{a_B}}}{z}\]
Here, rB is the radius of orbit, n is the energy level, aB is the Bohr's radius and z is the number of proton.
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 \[2.116 \times {10^{ - 10}}\;{\rm{m}}\].
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