A hydrogen like atom (described by the Bohr model) is observed ot emit ten wavelengths, originating from all possible transition between a group of levels. These levels have energies between - 0.85 eV and -0.544 eV (including both these values). (a) Find the atomic number of the atom.
(b) Calculate the smallest wavelength emitted in these transitions. (Take ground state energy of hydrogen atom =- 13.6 eV)

Explanation

To solve the problem step by step, we will break it down into two parts as requested.

### Part (a): Find the atomic number of the atom.

1. **Understanding the Problem**: We have a hydrogen-like atom that emits 10 wavelengths corresponding to transitions between energy levels. The energies of these levels are between -0.85 eV and -0.544 eV.

2. **Using the Formula for Transitions**: The number of transitions (or wavelengths emitted) between \( n \) energy levels is given by the formula:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]
We know that this equals 10. Therefore, we can set up the equation:
\[ \frac{n(n-1)}{2} = 10 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 20 \]

3. **Solving the Quadratic Equation**: Rearranging gives:
\[ n^2 - n - 20 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -20 \):
\[ n = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm 9}{2} \]
This gives us two possible solutions:
\[ n = 5 \quad \text{or} \quad n = -4 \]
Since \( n \) must be a positive integer, we have \( n = 5 \).

4. **Finding the Energy Levels**: The energy levels for a hydrogen-like atom are given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
We know the energies at the first and last levels:
\[ E_1 = -0.85 \text{ eV} \quad \text{and} \quad E_5 = -0.544 \text{ eV} \]

5. **Setting Up the Equations**:
For the first level:
\[ -0.85 = -\frac{13.6 Z^2}{1^2} \implies 0.85 = \frac{13.6 Z^2}{1} \implies Z^2 = \frac{0.85}{13.6} \]
For the fifth level:
\[ -0.544 = -\frac{13.6 Z^2}{5^2} \implies 0.544 = \frac{13.6 Z^2}{25} \implies Z^2 = \frac{0.544 \times 25}{13.6} \]

6. **Calculating \( Z \)**:
From the first equation:
\[ Z^2 = \frac{0.85}{13.6} \approx 0.0625 \implies Z \approx 0.25 \]
From the second equation:
\[ Z^2 = \frac{0.544 \times 25}{13.6} \approx 1 \]
Solving gives \( Z = 4 \).

### Part (b): Calculate the smallest wavelength emitted in these transitions.

1. **Finding the Energy Difference**: The smallest wavelength corresponds to the largest energy difference. The largest energy difference will be between the highest and lowest energy levels:
\[ \Delta E = E_1 - E_5 = -0.544 - (-0.85) = 0.306 \text{ eV} \]

2. **Using the Energy-Wavelength Relation**: The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
Where \( h \approx 4.1357 \times 10^{-15} \text{ eV s} \) and \( c \approx 3 \times 10^8 \text{ m/s} \).

3. **Calculating Wavelength**:
\[ \lambda = \frac{(4.1357 \times 10^{-15})(3 \times 10^8)}{0.306} \approx 4.052 \times 10^{-7} \text{ m} \approx 4052 \text{ nm} \]

### Final Answers:
- (a) The atomic number \( Z \) is 4.
- (b) The smallest wavelength emitted is approximately 4052 nm.