Hydrogen ($$_1H^1$$), Deuterium ($$_1H^2$$), singly ionised Helium ($$_2He^4$$)$$^+$$ and doubly ionised lithium ($$_3Li^6$$)$$^{++}$$ all have one electron around the nucleus. Consider an electron transition from $$n = 2$$ to $$n = 1$$. If the wave lengths of emitted radiation are $$\lambda_1$$, $$\lambda_2$$, $$\lambda_3$$ and $$\lambda_4$$ respectively then approximately which one of the following is correct?

For every hydrogen-like species the electron energy in the Bohr model is written first: $$E_n = -\,13.6\;\text{eV}\;\dfrac{Z^2}{n^2},$$ where $$Z$$ is the atomic number and $$n$$ is the principal quantum number. The transition energy from level $$n_2$$ to level $$n_1$$ is then $$\Delta E = E_{n_1}-E_{n_2} = -13.6\,\text{eV}\,Z^2\!\left(\dfrac{1}{n_1^{\,2}}-\dfrac{1}{n_2^{\,2}}\right).$$

Bohr also gives the wave-number formula $$\dfrac{1}{\lambda} = R\,Z^2\!\left(\dfrac{1}{n_1^{\,2}}-\dfrac{1}{n_2^{\,2}}\right),$$ where $$R$$ is the Rydberg constant. Since the present question asks only for ratios of wavelengths, we may use the proportionality $$\lambda \propto \dfrac{1}{Z^2}\;\;(\text{for given }n_1,n_2).$$ The difference between hydrogen and deuterium comes only from the reduced mass, which changes $$R$$ by about 0.05 %, far smaller than the integer ratios we are comparing, so we shall treat the two $$Z = 1$$ nuclei as giving the same wavelength.

We are concerned with the transition $$n_2 = 2 \rightarrow n_1 = 1$$, so the factor $$\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)=\dfrac{3}{4}$$ is the same for all four species. Therefore, denoting the wavelengths by $$\lambda_1$$ for $$_1H^1$$, $$\lambda_2$$ for $$_1H^2$$, $$\lambda_3$$ for $$_2He^4{}^+$$ and $$\lambda_4$$ for $$_3Li^6{}^{++}$$, we have

$$\lambda_1 \;:\;\lambda_2 \;:\;\lambda_3 \;:\;\lambda_4 \;=\;\dfrac{1}{Z_1^{\,2}}\;:\;\dfrac{1}{Z_2^{\,2}}\;:\;\dfrac{1}{Z_3^{\,2}}\;:\;\dfrac{1}{Z_4^{\,2}} \;=\;\dfrac{1}{1^{2}}\;:\;\dfrac{1}{1^{2}}\;:\;\dfrac{1}{2^{2}}\;:\;\dfrac{1}{3^{2}}.$$ Simplifying each denominator we obtain

$$\lambda_1 : \lambda_2 : \lambda_3 : \lambda_4 = 1 : 1 : \dfrac{1}{4} : \dfrac{1}{9}.$$

To express all four as integers, multiply every term by 36 (the LCM of 1, 4 and 9):

$$36\lambda_1 : 36\lambda_2 : 36\lambda_3 : 36\lambda_4 = 36 : 36 : 9 : 4.$$

Dividing each of these four numbers by 4 gives a cleaner ratio:

$$9\lambda_1 : 9\lambda_2 : 9\lambda_3 : 9\lambda_4 = 9 : 9 : 2.25 : 1.$$

It is clearer, however, to leave the ratio in its simplest fractional form, matching it directly to one of the given options:

$$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4.$$

Comparing with the options supplied in the problem statement, this equality is precisely the one written in Option C. Hence, the correct answer is Option C.