The radius of the hydrogen atom in its ground state is \( 5.3\times10^{-11}\,\mathrm{m} \). After a collision with an electron, it is found to have a radius of \( 21.2\times10^{-11}\,\mathrm{m} \). What is the principal quantum number n of the final state of the atom?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → n = 2
In the Bohr model the radius of the nth orbit is \( r_n=n^2a_0 \), where \( a_0 \) is the radius of the ground state (n = 1). So the radius grows as the square of the principal quantum number.
\( \begin{aligned} n^2 &= \frac{r_n}{a_0} \\ &= \frac{21.2\times10^{-11}\,\mathrm{m}}{5.3\times10^{-11}\,\mathrm{m}} \\ &= 4 \end{aligned} \)
\( \Rightarrow\; n=2 \)