If the lowest energy X-rays have `lambda=3.055xx10^(-8)` m, estimate the minimum difference in energy between two Bohr's orbits such that an electronic transition would correspond to the emission of an X-ray. Assuming that the electrons in other shells exert no influence, at what Z (minimum) would a transition form the second level to the first result in the emission of an X-ray?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Energy of the X-ray
We can use the formula for energy (E) of a photon, given by:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.63 \times 10^{-34} \) J·s (Planck's constant)
- \( c = 3.00 \times 10^{8} \) m/s (speed of light)
- \( \lambda = 3.055 \times 10^{-8} \) m (wavelength of the X-ray)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^{8})}{3.055 \times 10^{-8}} \]

Calculating this gives:
\[ E \approx 6.52 \times 10^{-18} \text{ J} \]

### Step 2: Calculate the Energy Difference Between Bohr's Orbits
For a hydrogen-like atom, the energy difference (\( \Delta E \)) between two energy levels can be calculated using:
\[ \Delta E = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R = 2.179 \times 10^{-18} \) J (Rydberg constant)
- \( n_1 = 1 \) (lower energy level)
- \( n_2 = 2 \) (higher energy level)

Substituting the values:
\[ \Delta E = 2.179 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ \Delta E = 2.179 \times 10^{-18} \left( 1 - \frac{1}{4} \right) \]
\[ \Delta E = 2.179 \times 10^{-18} \left( \frac{3}{4} \right) \]
\[ \Delta E \approx 1.63 \times 10^{-18} \text{ J} \]

### Step 3: Calculate the Minimum Atomic Number (Z)
To find the minimum Z such that the transition from the second level to the first results in the emission of an X-ray, we use the relationship:
\[ E = \Delta E \cdot Z^2 \]
Rearranging gives:
\[ Z^2 = \frac{E}{\Delta E} \]
Substituting the values we calculated:
\[ Z^2 = \frac{6.52 \times 10^{-18}}{1.63 \times 10^{-18}} \]
\[ Z^2 \approx 4 \]
Taking the square root:
\[ Z \approx 2 \]

### Conclusion
The minimum atomic number \( Z \) for which a transition from the second level to the first results in the emission of an X-ray is \( Z = 2 \), which corresponds to helium.

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