How many times larger is the spacing between the energy levels with n=3 and n=4,then the spacing between the energy levels with n=8 and n=9 for a hydrogen like atom or ion?

Options

A. 0.71
B. 0.41
C. 2.43
D. 14.82

Answer

To solve the problem, we need to find the difference in energy levels for a hydrogen-like atom or ion between the quantum states \( n=3 \) and \( n=4 \), and between \( n=8 \) and \( n=9 \). We can use the formula for the energy levels of a hydrogen-like atom:

\[ E_n = -\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n^2} \]

where:
- \( Z \) is the atomic number,
- \( k \) is Coulomb's constant,
- \( e \) is the charge of the electron,
- \( m \) is the mass of the electron,
- \( \hbar \) is the reduced Planck's constant,
- \( n \) is the principal quantum number.

### Step 1: Calculate the energy difference between \( n=3 \) and \( n=4 \)

The energy difference \( \Delta E_{34} \) can be calculated as:

\[ \Delta E_{34} = E_4 - E_3 = -\frac{Z^2 k e^4 m}{2 \hbar^2} \left( \frac{1}{4^2} - \frac{1}{3^2} \right) \]

Calculating \( \frac{1}{4^2} - \frac{1}{3^2} \):

\[ \frac{1}{16} - \frac{1}{9} = \frac{9 - 16}{144} = -\frac{7}{144} \]

Thus,

\[ \Delta E_{34} = -\frac{Z^2 k e^4 m}{2 \hbar^2} \left( -\frac{7}{144} \right) = \frac{7 Z^2 k e^4 m}{288 \hbar^2} \]

### Step 2: Calculate the energy difference between \( n=8 \) and \( n=9 \)

Now, we calculate \( \Delta E_{89} \):

\[ \Delta E_{89} = E_9 - E_8 = -\frac{Z^2 k e^4 m}{2 \hbar^2} \left( \frac{1}{9^2} - \frac{1}{8^2} \right) \]

Calculating \( \frac{1}{9^2} - \frac{1}{8^2} \):

\[ \frac{1}{81} - \frac{1}{64} = \frac{64 - 81}{5184} = -\frac{17}{5184} \]

Thus,

\[ \Delta E_{89} = -\frac{Z^2 k e^4 m}{2 \hbar^2} \left( -\frac{17}{5184} \right) = \frac{17 Z^2 k e^4 m}{10368 \hbar^2} \]

### Step 3: Find the ratio of the two energy differences

Now, we need to find how many times larger \( \Delta E_{34} \) is compared to \( \Delta E_{89} \):

\[ \text{Ratio} = \frac{\Delta E_{34}}{\Delta E_{89}} = \frac{\frac{7 Z^2 k e^4 m}{288 \hbar^2}}{\frac{17 Z^2 k e^4 m}{10368 \hbar^2}} \]

The \( Z^2 k e^4 m / \hbar^2 \) terms cancel out:

\[ \text{Ratio} = \frac{7}{288} \cdot \frac{10368}{17} = \frac{7 \cdot 36}{17} = \frac{252}{17} \approx 14.82 \]

### Conclusion

Thus, the spacing between the energy levels with \( n=3 \) and \( n=4 \) is approximately 14.82 times larger than the spacing between the energy levels with \( n=8 \) and \( n=9 \).