The ionization energy of a hydrogen like Bohr atom is `4` Rydberg. If the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state is `N-m` and if the radius of the first orbit of this atom is `r-m` then the value of `(N)/(r )=Pxx10^(2)` then, value of `P`. (Bohr radius of hydrogen `=5xx10^(-11)m,1` Rydberg `=2.2xx10^(-18)J`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Atomic Number (Z)
The ionization energy \( E_{\text{ionization}} \) is given as \( 4 \) Rydbergs. The energy of the nth orbit in a hydrogen-like atom is given by:
\[ E_n = -R_H \cdot \frac{Z^2}{n^2} \]
where \( R_H \) is the Rydberg constant for hydrogen, which is \( 1 \) Rydberg \( = 2.2 \times 10^{-18} \, J \).

For the ground state (n=1):
\[ E_1 = -R_H \cdot Z^2 \]
For ionization (n approaches infinity):
\[ E_{\infty} = 0 \]
Thus, the ionization energy can be expressed as:
\[ E_{\text{ionization}} = E_{\infty} - E_1 = 0 - (-R_H \cdot Z^2) = R_H \cdot Z^2 \]
Setting this equal to \( 4 \) Rydbergs:
\[ R_H \cdot Z^2 = 4 \cdot R_H \]
Cancelling \( R_H \):
\[ Z^2 = 4 \implies Z = 2 \]

### Step 2: Calculate the Wavelength (N)
The electron jumps from the first excited state (n=2) to the ground state (n=1). The energy difference between these states is:
\[ \Delta E = E_2 - E_1 = -R_H \cdot \frac{Z^2}{2^2} - (-R_H \cdot Z^2) = R_H \cdot Z^2 \left(1 - \frac{1}{4}\right) = R_H \cdot Z^2 \cdot \frac{3}{4} \]
Substituting \( Z = 2 \):
\[ \Delta E = R_H \cdot 4 \cdot \frac{3}{4} = 3 \cdot R_H \]
Using the relation \( E = \frac{hc}{\lambda} \):
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting \( \Delta E = 3 \cdot R_H \):
\[ \lambda = \frac{hc}{3 \cdot R_H} \]
Using \( h = 6.63 \times 10^{-34} \, J \cdot s \) and \( c = 3 \times 10^8 \, m/s \):
\[ \lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{3 \cdot (2.2 \times 10^{-18})} \]
Calculating:
\[ \lambda = \frac{1.989 \times 10^{-25}}{6.6 \times 10^{-18}} \approx 3.01 \times 10^{-8} \, m = 3 \, \text{nm} \]

### Step 3: Calculate the Radius (r)
The radius of the first orbit is given by:
\[ r_n = \frac{n^2 \cdot R_B}{Z} \]
For \( n = 1 \):
\[ r_1 = \frac{1^2 \cdot R_B}{Z} = \frac{R_B}{2} \]
Given \( R_B = 5 \times 10^{-11} \, m \):
\[ r_1 = \frac{5 \times 10^{-11}}{2} = 2.5 \times 10^{-11} \, m \]

### Step 4: Calculate \( \frac{N}{r} \)
Now we can find \( \frac{N}{r} \):
\[ \frac{N}{r} = \frac{3 \times 10^{-8}}{2.5 \times 10^{-11}} = 1.2 \times 10^3 \]
This can be expressed as:
\[ \frac{N}{r} = 1.2 \times 10^3 = P \times 10^2 \]
Thus, \( P = 12 \).

### Final Answer
The value of \( P \) is \( 12 \).

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