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Step 1: Understand the Context
We are given that hydrogen atoms are excited to a higher energy level and subsequently emit radiation of 6 different wavelengths. We need to determine the transition (between principal quantum numbers) that corresponds to the maximum wavelength among these emitted radiations.
Step 2: Number of Spectral Lines Formula
When an electron in a hydrogen atom is excited to the nth energy level and then allowed to de-excite, the total number of spectral lines emitted is given by the formula:
$$
\frac{n(n-1)}{2}.
$$
We are told this number is 6. So we set up the equation:
$$
\frac{n(n-1)}{2} = 6.
$$
Step 3: Solve for n
Multiply both sides by 2:
$$
n(n-1) = 12.
$$
Rewrite and rearrange as a standard quadratic equation:
$$
n^2 - n - 12 = 0.
$$
Factorize:
$$
(n - 4)(n + 3) = 0.
$$
From this, we get:
$$
n = 4 \quad \text{or} \quad n = -3.
$$
Since the principal quantum number n must be positive, we take
$$
n = 4.
$$
Step 4: Identify Possible Transitions
With the electron initially in the n=4 state, it can transition down to lower levels (n=3, n=2, or n=1) in different steps, emitting distinct wavelengths for each transition.
Step 5: Determine the Maximum Wavelength Transition
The energy of a photon emitted during a transition from n2 to n1 is:
$$
\Delta E = E_{n_2} - E_{n_1} \quad \Rightarrow \quad \Delta E \propto \frac{1}{\lambda}.
$$
Hence, a smaller energy difference corresponds to a larger wavelength, because
$$
\lambda \propto \frac{1}{\Delta E}.
$$
Among the allowed transitions from n=4 (i.e., 4→3, 4→2, 4→1, etc.), the smallest energy difference is from n=4 to n=3. Thus, the transition that provides the maximum wavelength is:
$$
n=4 \rightarrow n=3.
$$
