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
The wave number of electromagnetic radiation emitted during the transition of electron in between the two levels of `Li^(2+)` ion whose pricipal quantum numbner sum is `4` and difference is `2` is

Options

A. `3.5R_(H)`
B. `4R_(H)`
C. `8R_(H)`
D. `(8)/(9)R_(H)`

Explanation

To solve the problem of finding the wave number of electromagnetic radiation emitted during the transition of an electron in the Li²⁺ ion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The principal quantum number sum is 4.
- The difference between the principal quantum numbers is 2.
- The atomic number (Z) of lithium (Li) is 3.

2. **Determine the Principal Quantum Numbers (n1 and n2):**
- Let n1 be the lower energy level and n2 be the higher energy level.
- From the given sum (n1 + n2 = 4) and difference (n2 - n1 = 2), we can set up the equations:
- \( n1 + n2 = 4 \)
- \( n2 - n1 = 2 \)

- Solving these equations:
- From the second equation, we can express n2 as \( n2 = n1 + 2 \).
- Substitute this into the first equation:
\[ n1 + (n1 + 2) = 4 \implies 2n1 + 2 = 4 \implies 2n1 = 2 \implies n1 = 1 \]
- Now substituting \( n1 \) back to find \( n2 \):
\[ n2 = n1 + 2 = 1 + 2 = 3 \]

3. **Use the Rydberg Formula for Wave Number:**
- The formula for the wave number (ν̅) is given by:
\[ \bar{v} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R_H \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

4. **Substituting the Values:**
- Substitute \( Z = 3 \), \( n_1 = 1 \), and \( n_2 = 3 \) into the formula:
\[ \bar{v} = R_H \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ = R_H \cdot 9 \left( 1 - \frac{1}{9} \right) \]
\[ = R_H \cdot 9 \left( \frac{9 - 1}{9} \right) = R_H \cdot 9 \cdot \frac{8}{9} \]
\[ = 8 R_H \]

5. **Final Result:**
- The wave number of the electromagnetic radiation emitted during the transition is:
\[ \bar{v} = 8 R_H \]