The ionization energy of a hydrogen like Bohr atom is 5 Rydberg, (a) What is the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state?
(b) What is the radius of the first orbit for this atom? Given that Bohr radius of hydrogen atom = `5 xx 10^(-11) m` and 1 Rydberg = `2.2xx10^(-18)J`

Answer

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Wavelength of Radiation Emitted

1. **Calculate the Energy of the Ground State (E1)**:
The ionization energy of the hydrogen-like atom is given as 5 Rydbergs.
\[ E_1 = 5 \times 2.2 \times 10^{-18} \text{ J} = 11 \times 10^{-18} \text{ J} \]

2. **Calculate the Energy of the First Excited State (E2)**:
For a hydrogen-like atom, the energy of the nth state is given by:
\[ E_n = \frac{E_1}{n^2} \]
For the first excited state (n=2):
\[ E_2 = \frac{E_1}{2^2} = \frac{11 \times 10^{-18}}{4} = 2.75 \times 10^{-18} \text{ J} \]

3. **Calculate the Energy Difference (ΔE)**:
The energy emitted when the electron jumps from the first excited state to the ground state is:
\[ \Delta E = E_1 - E_2 = 11 \times 10^{-18} - 2.75 \times 10^{-18} = 8.25 \times 10^{-18} \text{ J} \]

4. **Use the Energy to Find the Wavelength (λ)**:
The energy of the emitted radiation is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting \(h = 6.626 \times 10^{-34} \text{ J s}\) and \(c = 3 \times 10^8 \text{ m/s}\):
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{8.25 \times 10^{-18}} \approx 2.4 \times 10^{-7} \text{ m} = 240 \text{ nm} \]

### Part (b): Radius of the First Orbit

1. **Use the Formula for the Radius of the First Orbit**:
The radius of the first orbit in a hydrogen-like atom is given by:
\[ r_1 = \frac{r_0}{Z} \]
where \(r_0\) is the Bohr radius of hydrogen and \(Z\) is the atomic number. Here, \(Z = 2\) for a hydrogen-like atom.

2. **Substitute the Values**:
Given \(r_0 = 5 \times 10^{-11} \text{ m}\):
\[ r_1 = \frac{5 \times 10^{-11}}{2} = 2.5 \times 10^{-11} \text{ m} \]

### Final Answers:
- (a) The wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state is **240 nm**.
- (b) The radius of the first orbit for this atom is **2.5 x 10^(-11) m**.