To solve the problem, we need to find the longest wavelength emitted in the radiation spectrum observed from hydrogen atoms transitioning between energy levels. Here’s a step-by-step breakdown of the solution:
### Step 1: Identify Energy Levels
- The ground state (A) corresponds to \( n = 1 \).
- The first excited state (B) corresponds to \( n = 2 \).
- The energy levels for hydrogen can be calculated using the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
- For \( n = 1 \):
\[
E_1 = -13.6 \, \text{eV}
\]
- For \( n = 2 \):
\[
E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV}
\]
- For \( n = 3 \):
\[
E_3 = -\frac{13.6}{3^2} \approx -1.51 \, \text{eV}
\]
- For \( n = 4 \):
\[
E_4 = -\frac{13.6}{4^2} = -0.85 \, \text{eV}
\]
### Step 2: Calculate Energy Level C
- The atoms absorb a photon of energy \( 2.7 \, \text{eV} \) and transition to a higher energy level C.
- The energy of level C can be calculated as:
\[
E_C = E_A + 2.7 \, \text{eV} = -13.6 \, \text{eV} + 2.7 \, \text{eV} = -10.9 \, \text{eV}
\]
- This energy level C corresponds to a value of \( n \) that is greater than 2.
### Step 3: Determine Possible Emission Transitions
- The atoms can emit photons of different energies as they transition back to lower energy levels.
- The possible transitions are:
- From C to B
- From C to A
- From B to A
- From C to E (if E is a higher level)
- From B to C
- From A to B
- The emitted photon energies will be the differences between the energy levels.
### Step 4: Find the Longest Wavelength
- The longest wavelength corresponds to the smallest energy difference (minimum energy gap).
- The energy differences can be calculated:
- \( E_{C} - E_{B} = -10.9 - (-3.4) = -7.5 \, \text{eV} \)
- \( E_{C} - E_{A} = -10.9 - (-13.6) = 2.7 \, \text{eV} \)
- \( E_{B} - E_{A} = -3.4 - (-13.6) = 10.2 \, \text{eV} \)
### Step 5: Calculate the Longest Wavelength
- The longest wavelength corresponds to the smallest energy gap, which is \( 2.7 \, \text{eV} \).
- The wavelength can be calculated using the formula:
\[
\lambda = \frac{hc}{E}
\]
where \( h = 6.626 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \).
- Convert \( 2.7 \, \text{eV} \) to joules:
\[
2.7 \, \text{eV} = 2.7 \times 1.6 \times 10^{-19} \, \text{J} = 4.32 \times 10^{-19} \, \text{J}
\]
- Substitute into the wavelength formula:
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{4.32 \times 10^{-19} \, \text{J}} \approx 4.6 \times 10^{-7} \, \text{m} = 4600 \, \text{nm}
\]
### Final Answer
The longest wavelength emitted in the radiation spectrum observed is approximately **4600 nm**.
