A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 $$\widetilde{A}$$. The radius of the atom in the excited state, in terms of Bohr radius $$a_0$$, will be:

For a hydrogen atom the energy of the level having principal quantum number $$n$$ is given by the Bohr formula

$$E_n=-\dfrac{13.6\ \text{eV}}{n^2}.$$

The atom is initially in its ground state, so its initial energy is

$$E_1=-13.6\ \text{eV}.$$

When it absorbs a photon of wavelength $$\lambda=980\ \text{\AA},$$ the energy of that photon is obtained from the relation $$E=\dfrac{hc}{\lambda}.$$

Using the useful constant $$hc\approx12400\ \text{eV\;\AA},$$ we have

$$E=\dfrac{12400\ \text{eV\;\AA}}{980\ \text{\AA}}=12.653\ \text{eV}\;(\text{approximately}).$$

This entire photon energy raises the electron from $$n=1$$ to some higher level $$n=n_f$$, so the energy conservation condition is

$$E=E_{n_f}-E_1.$$

Substituting the Bohr energies we obtain

$$12.653= \left(-\dfrac{13.6}{n_f^{2}}\right)-(-13.6)=13.6\left(1-\dfrac{1}{n_f^{2}}\right).$$

Dividing both sides by $$13.6$$ gives

$$1-\dfrac{1}{n_f^{2}}=\dfrac{12.653}{13.6}=0.9301.$$

Hence

$$\dfrac{1}{n_f^{2}}=1-0.9301=0.0699,$$

and therefore

$$n_f^{2}=\dfrac{1}{0.0699}\approx14.3.$$

The nearest integral value satisfying this is $$n_f=4$$ (because $$4^{2}=16$$ gives almost the same energy, whereas $$n=3$$ would be far off).

Thus the electron is excited to the fourth orbit.

For the Bohr model the radius of the $$n^{\text{th}}$$ orbit is

$$r_n=n^{2}a_{0},$$

so for $$n=4$$ we have

$$r_4=4^{2}a_{0}=16a_{0}.$$

Hence, the correct answer is Option C.