In the hydrogen atom spectrum, the least energetic photon takes place during the transition from n = 6 energy level to n = ……… energy level.

Options

A. 1
B. 3
C. 5
D. 4

Answer

To find the least energetic photon transition in the hydrogen atom spectrum from the n = 6 energy level, we can use the formula for the energy difference between two energy levels:

\[ \Delta E = E_n = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

where:
- \( n_i \) is the initial energy level (in this case, \( n_i = 6 \))
- \( n_f \) is the final energy level
- \( \Delta E \) is the energy of the photon emitted or absorbed during the transition.

### Step 1: Identify the initial energy level
The initial energy level is given as \( n_i = 6 \).

### Step 2: Determine the condition for the least energetic photon
To find the least energetic photon, we need to minimize \( \Delta E \). This occurs when \( n_f \) is as high as possible, but still less than \( n_i \). Therefore, \( n_f \) must be less than 6.

### Step 3: List possible values for \( n_f \)
The possible values for \( n_f \) that are less than 6 are:
- \( n_f = 5 \)
- \( n_f = 4 \)
- \( n_f = 3 \)
- \( n_f = 2 \)
- \( n_f = 1 \)

### Step 4: Calculate \( \Delta E \) for each possible \( n_f \)
We will calculate \( \Delta E \) for \( n_f = 5, 4, 3, 2, 1 \):

1. For \( n_f = 5 \):
\[ \Delta E = 2.18 \times 10^{-18} \left( \frac{1}{5^2} - \frac{1}{6^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{25} - \frac{1}{36} \right) \]

2. For \( n_f = 4 \):
\[ \Delta E = 2.18 \times 10^{-18} \left( \frac{1}{4^2} - \frac{1}{6^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{16} - \frac{1}{36} \right) \]

3. For \( n_f = 3 \):
\[ \Delta E = 2.18 \times 10^{-18} \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{9} - \frac{1}{36} \right) \]

4. For \( n_f = 2 \):
\[ \Delta E = 2.18 \times 10^{-18} \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{4} - \frac{1}{36} \right) \]

5. For \( n_f = 1 \):
\[ \Delta E = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{6^2} \right) = 2.18 \times 10^{-18} \left( 1 - \frac{1}{36} \right) \]

### Step 5: Compare the values of \( \Delta E \)
The least energetic transition will occur when \( n_f \) is the highest possible value, which is \( n_f = 5 \).

### Conclusion
Thus, the least energetic photon transition from \( n = 6 \) occurs when transitioning to \( n = 5 \).

**Final Answer: \( n = 5 \)**