Hydrogen Atom and Hydrogen Molecule
The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is
`1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5`
`R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1)`
Here,
`R_(H)` is the Rydberg Constant, `m_(e)` is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion.
From the given data, calculate the energy change for the process
`H^(-) rarr H+e^(-)`

Answer

To calculate the energy change for the process \( H^- \rightarrow H + e^- \), we can follow these steps:

### Step 1: Understanding the Process
The process involves the conversion of a hydrogen anion (\( H^- \)) into a neutral hydrogen atom (\( H \)) and an electron (\( e^- \)). The energy change associated with this process can be determined by calculating the energy of the photon emitted when the electron transitions from a higher energy level to a lower one.

### Step 2: Using Balmer's Formula
We will use Balmer's empirical formula to find the wavelength of the emitted photon. The formula is given by:

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

Where:
- \( R_H = 109678 \, \text{cm}^{-1} \) (Rydberg constant)
- \( n \) is the principal quantum number of the higher energy level.

For the transition from \( n = 3 \) to \( n = 2 \) (the first line of the Balmer series):

### Step 3: Calculate \( \frac{1}{\lambda} \)
Substituting \( n = 3 \) into the formula:

\[ \frac{1}{\lambda} = 109678 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the terms:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

Now substituting these values:

\[ \frac{1}{\lambda} = 109678 \left( 0.25 - 0.1111 \right) = 109678 \left( 0.1389 \right) \]

Calculating:

\[ \frac{1}{\lambda} \approx 15263.3 \, \text{cm}^{-1} \]

### Step 4: Calculate \( \lambda \)
Now, taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{15263.3} \approx 6.55 \times 10^{-5} \, \text{cm} = 6550 \, \text{Å} \]

### Step 5: Calculate the Energy of the Photon
The energy of the photon can be calculated using the formula:

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

Where:
- \( h = 4.135667696 \times 10^{-15} \, \text{eV s} \) (Planck's constant)
- \( c = 3 \times 10^{10} \, \text{cm/s} \) (speed of light)

Substituting the values:

\[ E = \frac{(4.135667696 \times 10^{-15} \, \text{eV s})(3 \times 10^{10} \, \text{cm/s})}{6550 \times 10^{-8} \, \text{cm}} \]

Calculating:

\[ E \approx \frac{1.240 \times 10^{-4} \, \text{eV cm}}{6.55 \times 10^{-5} \, \text{cm}} \approx 1.89 \, \text{eV} \]

### Step 6: Conclusion
The energy change for the process \( H^- \rightarrow H + e^- \) is approximately \( 1.89 \, \text{eV} \).

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