Estimate the difference in energy between 1st and 2nd Bohr orbits for hydrogen atom. At what minimum atomic number, a transition from n = 2 to n=1 energy level would result in the emission of X-rays with `lambda=3 xx 10^(-8) m? ` Which hydrogen atom like species does this atomic number correspond to?

Explanation

To solve the problem, we will break it down into two parts:

1. Estimating the difference in energy between the 1st and 2nd Bohr orbits for a hydrogen atom.
2. Finding the minimum atomic number for a transition from n = 2 to n = 1 that results in the emission of X-rays with a given wavelength.

### Part 1: Energy Difference Between 1st and 2nd Bohr Orbits

**Step 1: Energy Levels in the Bohr Model**
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{R_H}{n^2} \]
where \( R_H \) is the Rydberg constant for hydrogen, approximately \( 2.18 \times 10^{-18} \) Joules.

**Step 2: Calculate Energy for n = 1 and n = 2**
- For \( n = 1 \):
\[ E_1 = -\frac{R_H}{1^2} = -2.18 \times 10^{-18} \text{ J} \]
- For \( n = 2 \):
\[ E_2 = -\frac{R_H}{2^2} = -\frac{2.18 \times 10^{-18}}{4} = -0.545 \times 10^{-18} \text{ J} \]

**Step 3: Calculate the Energy Difference**
The difference in energy \( \Delta E \) between the first and second orbits is:
\[ \Delta E = E_2 - E_1 = -0.545 \times 10^{-18} - (-2.18 \times 10^{-18}) = 1.635 \times 10^{-18} \text{ J} \]

### Part 2: Minimum Atomic Number for X-ray Emission

**Step 4: Use the Wavelength to Find Energy**
The energy of a photon can also be calculated using the wavelength \( \lambda \):
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)). Given \( \lambda = 3 \times 10^{-8} \text{ m} \):
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3 \times 10^{-8}} = 6.626 \times 10^{-18} \text{ J} \]

**Step 5: Relate Energy to Atomic Number**
For hydrogen-like species, the energy difference can be expressed as:
\[ \Delta E = Z^2 \cdot R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \Delta E = Z^2 \cdot R_H \left( 1 - \frac{1}{4} \right) = Z^2 \cdot R_H \cdot \frac{3}{4} \]
Setting this equal to the energy calculated from the wavelength:
\[ Z^2 \cdot R_H \cdot \frac{3}{4} = 6.626 \times 10^{-18} \]

**Step 6: Solve for Z**
Substituting \( R_H = 2.18 \times 10^{-18} \):
\[ Z^2 \cdot \frac{3}{4} \cdot 2.18 \times 10^{-18} = 6.626 \times 10^{-18} \]
\[ Z^2 \cdot \frac{3 \cdot 2.18}{4} = 6.626 \]
\[ Z^2 = \frac{6.626 \cdot 4}{3 \cdot 2.18} \]
Calculating this gives:
\[ Z^2 \approx 4 \implies Z \approx 2 \]

### Conclusion
The minimum atomic number \( Z \) is 2, which corresponds to the hydrogen-like species \( \text{He}^+ \) (helium ion).