Hydrogen atom:
The electronic ground state of hydrogen atom contains one electron in the first orbit. If sufficient energy is provided, this electron can be promoted to higher energy levels. The electronic energy of a hydrogen-like species (any atom//ions with nuclear charge Z and one electron) can be given as
`E_(n)=-(R_(H)Z^(2))/(n^(2))` where `R_(H)= "Rydberg constant," n= "principal quantum number"`
Calculate the wavelength of light (nm) for the electronic transition of H-atom from the first excited state to ground state.
In the model of hydrogen like atom put forward by Niels Bohr `(1913)` the electron orbits around the central nucleus. the bohr radius of `n^(th)` orbit of a hydrogen-like species is given by
`r= k n^(2)/Z` " " where, k is constant

Answer

To calculate the wavelength of light for the electronic transition of the hydrogen atom from the first excited state to the ground state, we will follow these steps:

### Step 1: Identify the energy levels
The energy of the electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{R_H Z^2}{n^2} \]
For hydrogen, \( Z = 1 \).

### Step 2: Calculate the energy of the ground state (n=1)
Using \( n = 1 \):
\[ E_1 = -\frac{R_H \cdot 1^2}{1^2} = -R_H \]

### Step 3: Calculate the energy of the first excited state (n=2)
Using \( n = 2 \):
\[ E_2 = -\frac{R_H \cdot 1^2}{2^2} = -\frac{R_H}{4} \]

### Step 4: Calculate the energy difference between the two states
The energy difference (\( \Delta E \)) between the first excited state and the ground state is:
\[ \Delta E = E_2 - E_1 \]
\[ \Delta E = \left(-\frac{R_H}{4}\right) - \left(-R_H\right) \]
\[ \Delta E = -\frac{R_H}{4} + R_H \]
\[ \Delta E = R_H - \frac{R_H}{4} = \frac{3R_H}{4} \]

### Step 5: Use the energy-wavelength relationship
The energy of a photon is related to its wavelength (\( \lambda \)) by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \))

Rearranging for \( \lambda \):
\[ \lambda = \frac{hc}{\Delta E} \]

### Step 6: Substitute \( \Delta E \) into the wavelength equation
Substituting \( \Delta E = \frac{3R_H}{4} \):
\[ \lambda = \frac{hc}{\frac{3R_H}{4}} = \frac{4hc}{3R_H} \]

### Step 7: Calculate the wavelength
Using the value of \( R_H \) (approximately \( 2.18 \times 10^{-18} \, \text{J} \)):
\[ \lambda = \frac{4 \cdot (6.626 \times 10^{-34} \, \text{Js}) \cdot (3.00 \times 10^8 \, \text{m/s})}{3 \cdot (2.18 \times 10^{-18} \, \text{J})} \]

Calculating this gives:
\[ \lambda \approx \frac{7.957 \times 10^{-25}}{6.54 \times 10^{-18}} \]
\[ \lambda \approx 121.6 \, \text{nm} \]

### Final Answer:
The wavelength of light for the electronic transition of the H-atom from the first excited state to the ground state is approximately **121.6 nm**.