The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively
Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to `(n(n - 1))/(2)`If teh electron comes back from the energy level having energy `E_(2)` to the energy level having energy `E_(1)` then the difference may be expresent in terms of energy of photon as `E_(2) - E_(1) = Delta E, lambda = hc//Delta E` Since h and c are constants `Delta E` coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula `bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2)))`Where R is a Rydherg constant
If the ionisation potential for hydrogen -like atom in a sample is ` 122.4 V` then the series limit of the paschen series for this atom is

Options

A. R
B. `(R )/(3^(2))`
C. `(3^(2)R)/(4^(2))`
D. `3 ^(2)R`

Explanation

To find the series limit of the Paschen series for a hydrogen-like atom with an ionization potential of 122.4 V, we can follow these steps:

### Step 1: Understand the Ionization Energy
The ionization potential (or ionization energy) for hydrogen-like atoms is given as 122.4 eV. This value represents the energy required to remove an electron from the ground state of the atom.

### Step 2: Relate Ionization Energy to Atomic Number (Z)
The ionization energy for hydrogen-like atoms can be expressed using the formula:
\[ E = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( E \) is the ionization energy, \( Z \) is the atomic number, and \( n \) is the principal quantum number of the electron in the ground state (for hydrogen, \( n = 1 \)).

Given that the ionization energy is 122.4 eV, we can set up the equation:
\[ 122.4 = \frac{13.6 \cdot Z^2}{1^2} \]

### Step 3: Solve for Z
Rearranging the equation to solve for \( Z^2 \):
\[ Z^2 = \frac{122.4}{13.6} \]
Calculating this gives:
\[ Z^2 = 9 \]
Taking the square root:
\[ Z = 3 \]

### Step 4: Determine the Series Limit of the Paschen Series
The Paschen series corresponds to transitions where the electron falls from higher energy levels (n = 4, 5, ...) to the n = 3 level. The series limit occurs when the electron transitions from n = ∞ to n = 3.

Using the Rydberg formula for the wavelength:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = \infty \)

Thus, we can write:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{3^2} - 0 \right) \]
\[ \frac{1}{\lambda} = RZ^2 \cdot \frac{1}{9} \]

### Step 5: Substitute Z into the Formula
Substituting \( Z = 3 \):
\[ \frac{1}{\lambda} = R \cdot 3^2 \cdot \frac{1}{9} \]
\[ \frac{1}{\lambda} = R \]

### Conclusion
The series limit of the Paschen series for this hydrogen-like atom is equal to the Rydberg constant \( R \).