Calculate the energy emitted when electron of 1.0 gm atom of Hydrogen undergo transition giving the spectrtal lines of lowest energy is visible region of its atomic spectra. Given that, `R_(H)`=`1.1xx10^(7) m^(-1)`,` c=3xx10^8m//sec`,` h=6.625xx10^(-34) Jsec`.

Explanation

To calculate the energy emitted when an electron in a hydrogen atom undergoes a transition that gives rise to the spectral lines in the visible region, we will follow these steps:

### Step 1: Identify the Transition
The lowest energy transition in the visible region for hydrogen occurs in the Balmer series, specifically from the third energy level (n2 = 3) to the second energy level (n1 = 2).

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of the emitted light is given by:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting \( n_1 = 2 \) and \( n_2 = 3 \):

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]

Calculating the right-hand side:

\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

Thus,

\[ \frac{1}{\lambda} = R_H \cdot \frac{5}{36} \]

### Step 3: Substitute the Rydberg Constant
Given \( R_H = 1.1 \times 10^7 \, \text{m}^{-1} \):

\[ \frac{1}{\lambda} = 1.1 \times 10^7 \cdot \frac{5}{36} \]

Calculating \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{5.5 \times 10^7}{36} \approx 1.52778 \times 10^6 \, \text{m}^{-1} \]

### Step 4: Calculate the Wavelength (λ)
Taking the reciprocal gives us the wavelength:

\[ \lambda = \frac{1}{1.52778 \times 10^6} \approx 6.54 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate the Energy (E)
Using the formula for energy:

\[ E = \frac{hc}{\lambda} \]

Substituting \( h = 6.625 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):

\[ E = \frac{(6.625 \times 10^{-34}) \cdot (3 \times 10^8)}{6.54 \times 10^{-7}} \]

Calculating the energy:

\[ E \approx \frac{1.9875 \times 10^{-25}}{6.54 \times 10^{-7}} \approx 3.03 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate the Energy for 1 Gram of Hydrogen
To find the total energy emitted for 1 gram of hydrogen, we need to multiply the energy per atom by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \)):

\[ E_{\text{total}} = E \times N_A = (3.03 \times 10^{-19} \, \text{J}) \times (6.022 \times 10^{23} \, \text{atoms/mol}) \]

Calculating the total energy:

\[ E_{\text{total}} \approx 18.25 \, \text{J} \]

### Final Answer
The energy emitted when an electron of 1.0 gram of hydrogen undergoes the transition is approximately **18.25 J**.

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