A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelength of incident radiation should be

Options

A. 800 A
B. 825 A
C. 970 A
D. 1025 A

Answer

To solve the problem of determining the wavelength of incident radiation needed to produce six lines in the emission spectrum of a hydrogen atom in its ground state, we can follow these steps:

### Step 1: Understand the emission lines
To produce six lines in the emission spectrum, we need to determine how many energy levels the electron can transition between. The number of lines (L) produced when an electron transitions from a higher energy level to a lower one can be calculated using the formula:
\[ L = \frac{n(n-1)}{2} \]
where \( n \) is the number of energy levels involved.

### Step 2: Calculate the required energy levels
We need to find \( n \) such that:
\[ \frac{n(n-1)}{2} = 6 \]
Solving for \( n \):
\[ n(n-1) = 12 \]
Testing values:
- For \( n = 4 \): \( 4(4-1) = 12 \) (This works)
Thus, the electron must transition between 4 energy levels (from n=1 to n=4).

### Step 3: Determine energy levels
The energy of an electron in the nth orbit of a hydrogen atom is given by:
\[ E_n = -\frac{Z^2 \cdot E_0}{n^2} \]
where \( Z \) is the atomic number (1 for hydrogen) and \( E_0 = 13.6 \, \text{eV} \).

Calculating the energies for n=1 and n=4:
- For \( n=1 \):
\[ E_1 = -\frac{1^2 \cdot 13.6}{1^2} = -13.6 \, \text{eV} \]
- For \( n=4 \):
\[ E_4 = -\frac{1^2 \cdot 13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \]

### Step 4: Calculate the energy difference
Now, we find the energy difference \( \Delta E \) when the electron transitions from n=4 to n=1:
\[ \Delta E = E_1 - E_4 = (-13.6) - (-0.85) = -13.6 + 0.85 = -12.75 \, \text{eV} \]

### Step 5: Calculate the wavelength
Using the energy-wavelength relationship:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \, \text{eV s} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), we can rearrange to find \( \lambda \):
\[ \lambda = \frac{hc}{E} \]

Substituting the values:
\[ \lambda = \frac{(4.1357 \times 10^{-15} \, \text{eV s}) \cdot (3 \times 10^8 \, \text{m/s})}{12.75} \]

Calculating \( \lambda \):
\[ \lambda \approx \frac{1.240 \times 10^{-6} \, \text{eV m}}{12.75} \approx 9.70 \times 10^{-8} \, \text{m} \]
Converting to Angstroms (1 m = \( 10^{10} \) Angstroms):
\[ \lambda \approx 970 \, \text{Å} \]

### Conclusion
The wavelength of the incident radiation required to produce six lines in the emission spectrum of a hydrogen atom in its ground state is approximately **970 Å**.

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