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.
A formula analogous to Balmer's formula applies to the series of spectral lines which arise from transition from higher energy levels to the lowest energy level of hydrogen atom. Write this formula and use it to determine the ground state energy of a hydrogen atom in eV.
A 'muonic hydrogen atom' is like a hydrogen atom in which the electron is replaced by a heavier particle, the muon. The mass of a muon is about `207` times the mass of an electron, while its charge is the same as that of an electron. A muon has a very short lifetime, but we ignore its ubstable nature here.

Answer

To solve the question regarding the ground state energy of a hydrogen atom and to derive the analogous formula for the spectral lines, we will follow these steps:

### Step 1: Write the formula for the spectral lines of hydrogen
The formula for the wavelengths of the spectral lines in the Balmer series of hydrogen is given by:

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

where:
- \( R_H \) is the Rydberg constant, \( R_H = 109.678 \, \text{cm}^{-1} \)
- \( n \) is the principal quantum number, with \( n = 3, 4, 5, \ldots \)

### Step 2: Determine the ground state energy of the hydrogen atom
The ground state energy of a hydrogen atom can be derived from the Bohr model. The energy levels of a hydrogen atom are given by:

\[ E_n = -\frac{R_H \cdot h \cdot c}{n^2} \]

For the ground state, \( n = 1 \):

\[ E_1 = -\frac{R_H \cdot h \cdot c}{1^2} \]

### Step 3: Substitute the values to find \( E_1 \)
We need to convert the Rydberg constant from \( \text{cm}^{-1} \) to energy in electron volts (eV). The conversion factor is:

\[ 1 \, \text{cm}^{-1} = 1.23984 \times 10^{-4} \, \text{eV} \]

Thus,

\[ E_1 = -R_H \cdot h \cdot c \]

Substituting the values:

- \( R_H = 109678 \, \text{cm}^{-1} \)
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^{10} \, \text{cm/s} \)

Calculating \( E_1 \):

\[ E_1 = -109678 \times 6.626 \times 10^{-34} \times 3 \times 10^{10} \]

Calculating this gives:

\[ E_1 \approx -13.6 \, \text{eV} \]

### Final Result
The ground state energy of a hydrogen atom is approximately:

\[ E_1 \approx -13.6 \, \text{eV} \]

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