According to Bohr's Theory the wavelength of the radiation emitted from hydrogen atom is given by

Given information:
- The energy required to ionize a hydrogen-like ion in its ground state is 9 Rydbergs.
- The electron jumps from the second excited state to the ground state.

Explanation:
To solve this problem, we need to understand the concept of energy levels in hydrogen-like ions and the relationship between energy and wavelength.

Energy levels in hydrogen-like ions:
Hydrogen-like ions are ions that have a single electron orbiting a nucleus with a positive charge. The energy levels in these ions are given by the formula:

E = - (13.6 Z^2) / n^2

Where:
- E is the energy of the electron in the hydrogen-like ion
- Z is the atomic number of the nucleus
- n is the principal quantum number, which represents the energy level of the electron

The ground state of a hydrogen-like ion corresponds to n = 1, and the energy of the electron in the ground state is given by:

E1 = - (13.6 Z^2)

The energy required to ionize the hydrogen-like ion in its ground state is the energy difference between the ground state and when the electron is completely removed, which is given by:

ΔE = E∞ - E1

Where:
- ΔE is the energy required for ionization
- E∞ is the energy of the electron when it is completely removed from the ion

In this case, the energy required to ionize the hydrogen-like ion in its ground state is given as 9 Rydbergs.

Relationship between energy and wavelength:
The relationship between energy and wavelength is given by the equation:

E = hc / λ

Where:
- E is the energy of the radiation
- h is the Planck's constant (6.626 x 10^-34 J s)
- c is the speed of light (3.0 x 10^8 m/s)
- λ is the wavelength of the radiation

Calculating the wavelength:
To calculate the wavelength of the radiation emitted when the electron jumps from the second excited state to the ground state, we need to find the energy difference between these two states and then use the energy-wavelength relationship.

The energy of the electron in the second excited state is given by:

E2 = - (13.6 Z^2) / 4^2

The energy difference between the second excited state and the ground state is given by:

ΔE = E2 - E1

Substituting the values:

ΔE = - (13.6 Z^2) / 4^2 - (- (13.6 Z^2))

Simplifying the expression:

ΔE = (13.6 Z^2) - (13.6 Z^2) / 16

ΔE = (15/16) * (13.6 Z^2)

Given that ΔE = 9 Rydbergs, we can solve for Z:

9 = (15/16) * (13.6 Z^2)

Z^2 = (9 * 16) / (15 * 13.6)

Z^2 = 0.7058

Z ≈ 0.840

Now, we can calculate the wavelength using the energy-wavelength relationship:

E = hc / λ

9 = (6