$$X$$ different wavelength may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number $$n = 6$$? The value of $$X$$ is _________.

In a hydrogen atom each spectral line arises when the electron falls from an upper level with principal quantum number $$n_i$$ to a lower level $$n_f$$, where $$n_i > n_f$$. Because the energy difference $$\Delta E$$ and, therefore, the wavelength $$\lambda$$ of the emitted photon depend only on these two numbers, every different ordered pair $$(n_i,n_f)$$ gives a unique wavelength.

The atoms in the sample are excited only up to the level $$n_{\text{max}} = 6$$. Hence the possible values of $$n_i$$ are $$6,5,4,3,2$$ (the ground state $$n=1$$ cannot serve as an upper level). For each chosen $$n_i$$, the lower level $$n_f$$ can be any integer from $$1$$ up to $$(n_i-1)$$.

We count the number of different pairs $$(n_i,n_f)$$ by listing them:

For $$n_i = 6: \; n_f = 5,4,3,2,1 \;\Rightarrow\; 5 \text{ lines}$$

For $$n_i = 5: \; n_f = 4,3,2,1 \;\Rightarrow\; 4 \text{ lines}$$

For $$n_i = 4: \; n_f = 3,2,1 \;\Rightarrow\; 3 \text{ lines}$$

For $$n_i = 3: \; n_f = 2,1 \;\Rightarrow\; 2 \text{ lines}$$

For $$n_i = 2: \; n_f = 1 \;\Rightarrow\; 1 \text{ line}$$

Adding all these possibilities, we obtain

$$X = 5 + 4 + 3 + 2 + 1 = 15.$$

This result is often written with the combination formula

$$X = \frac{n_{\text{max}}\,(n_{\text{max}}-1)}{2} = \frac{6 \times 5}{2} = 15,$$

because the total number of distinct pairs among $$n_{\text{max}}$$ levels is the number of ways to choose two different levels, and that is $$\frac{n(n-1)}{2}.$$

So, the answer is $$15$$.