A hydrogen like atom (atomic number Z) is in a higher excited satte of quantum number n .This excited atom can make a transition to the first excited state by succesively emitting two photon of energies `10.20 eV` and `17.00 eV` .Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting twio photon of energy `4.25 ev` and `5.95 eV` Determine the followings:
The atom during transition from `n = 1` to `n = 2` emit radiation in the region of

Options

A. Visible
B. Infira-red
C. UV
D. None

Explanation

To solve the problem step by step, we need to determine the wavelength of the radiation emitted during the transition from \( n = 1 \) to \( n = 2 \) for a hydrogen-like atom with atomic number \( Z \).

### Step 1: Understand the Energy Levels
For a hydrogen-like atom, the energy levels are given by the formula:
\[ E_n = -\frac{Z^2 \cdot R}{n^2} \]
where \( R \) is the Rydberg constant, approximately \( 13.6 \, \text{eV} \) for hydrogen.

### Step 2: Calculate the Energy Difference
To find the energy difference between the levels \( n = 1 \) and \( n = 2 \):
\[ E_{1} = -\frac{Z^2 \cdot R}{1^2} = -Z^2 \cdot R \]
\[ E_{2} = -\frac{Z^2 \cdot R}{2^2} = -\frac{Z^2 \cdot R}{4} \]
The energy difference \( \Delta E \) when transitioning from \( n = 1 \) to \( n = 2 \) is:
\[ \Delta E = E_{2} - E_{1} = \left(-\frac{Z^2 \cdot R}{4}\right) - \left(-Z^2 \cdot R\right) = Z^2 \cdot R \left(1 - \frac{1}{4}\right) = Z^2 \cdot R \cdot \frac{3}{4} \]

### Step 3: Calculate the Wavelength
Using the energy-wavelength relationship:
\[ E = \frac{hc}{\lambda} \]
we can rearrange this to find the wavelength \( \lambda \):
\[ \lambda = \frac{hc}{E} \]
Substituting \( E = Z^2 \cdot R \cdot \frac{3}{4} \):
\[ \lambda = \frac{hc}{Z^2 \cdot R \cdot \frac{3}{4}} = \frac{4hc}{3Z^2 \cdot R} \]

### Step 4: Substitute Values
Using \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \), and \( R = 13.6 \, \text{eV} \) (which we need to convert to Joules by multiplying by \( 1.6 \times 10^{-19} \)):
\[ R = 13.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.176 \times 10^{-18} \, \text{J} \]
Now substituting these values into the wavelength equation:
\[ \lambda = \frac{4 \cdot (6.626 \times 10^{-34}) \cdot (3 \times 10^8)}{3Z^2 \cdot (2.176 \times 10^{-18})} \]

### Step 5: Simplify the Expression
Calculating the constants:
\[ \lambda = \frac{7.9512 \times 10^{-25}}{6.528 \times 10^{-18}Z^2} = \frac{1.216 \times 10^{-7}}{Z^2} \]

### Step 6: Determine the Region of Wavelength
The wavelength \( \lambda \) can be converted to nanometers (1 nm = \( 10^{-9} \) m):
\[ \lambda = \frac{1.216 \times 10^{-7}}{Z^2} \times 10^9 = \frac{121.6}{Z^2} \, \text{nm} \]

### Conclusion
The emitted radiation during the transition from \( n = 1 \) to \( n = 2 \) will be in the ultraviolet region for \( Z > 1 \) since the wavelength will be less than 400 nm.