Question

The radius of H-atom in its ground state is $5.3\times {10}^{-11}$m. After a collision with an electron, it is found to have a radius of $21.2\times {10}^{-11}$m. What is the principal quantum number 'n' of the final state of the atom?

• A. $n=2$
• B. $n=3$
• C. $n=4$
• D. $n=16$

Answer 1

Answer: (A)

$n=2$

Let us consider the problem:

Find the principle quantum number of the final state of the atom.

$\frac{r_1}{r_2}=\frac{n_1^2}{n_2^2}$

Implies that

$\frac{1}{n_2^2}=\frac{5.3\times {10}^{-11}}{21.2\times {10}^{-11}}$

Implies that

$n_2^2=4$

Hence,

$n_2=2$.