Question

Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda$ = 975 $\mathring{A}$. Number of spectral lines in the resulting spectrum emitted will be

• A. 3
• B. 2
• C. 6
• D. 10

Answer 1

The Correct Option is C

To determine the number of spectral lines emitted by a hydrogen atom, when it is excited by a monochromatic radiation, we follow these steps:

1. First, calculate the energy of the monochromatic radiation which excites the hydrogen atom. The energy of the radiation can be calculated using the wavelength (\(\lambda\)) with the formula:

   E = \frac{hc}{\lambda}

   Where:

   • h = 6.626 \times 10^{-34} \, J \cdot s (Planck's constant)
   • c = 3 \times 10^8 \, m/s (speed of light)
   • \lambda = 975 \, \mathring{A} = 975 \times 10^{-10} \, m

2. Substituting the given values, we get:

   E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{975 \times 10^{-10}} = \frac{19.878 \times 10^{-26}}{9.75 \times 10^{-8}} = 20.387 \times 10^{-19} \, J \approx 2.09 \, eV

3. Find the energy levels of the hydrogen atom using Bohr's model. The energy of the nth level is given by:

   E_n = - 13.6 \, \text{eV} \cdot \frac{1}{n^2}

4. Determine from which level the electron was excited to, using the calculated energy:

   E_{ground} = -13.6 \, eV and E = E_{n} - E_{ground}

   Thus, 2.09 = \frac{13.6}{n^2}

   Simplifying gives:

   n^2 = \frac{13.6}{2.09} \approx 6.5

   The nearest whole number for \(n\) is 3, suggesting the electron was excited from ground state (n=1) to n=3.

5. Calculate the number of spectral lines formed when the electron transitions between these levels. The number of lines is given by:

   N = \frac{n(n-1)}{2}

   Substitute \(n=4\) because it includes all possible transitions: 2 to 3 and 3 to 1.

   N = \frac{4(4-1)}{2} = \frac{12}{2} = 6

Therefore, the correct answer is 6.