Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be :

Question

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Answer 1

To solve this problem, we need to use Bohr's theory of the hydrogen atom, which explains the quantized nature of electron energy levels.

The ionization potential of hydrogen is given as 13.6 eV, which means an electron in the ground state (n = 1) has an energy of -13.6 eV. When a hydrogen atom absorbs a photon of energy 12.1 eV, the electron transitions from the ground state to a higher energy level. Start by calculating the new energy level to which the electron is excited.

The energy absorbed by the electron can be written as:

E = 13.6 - E_n

Given:

E = 12.1 \text{ eV}

The energy of the electron in the nth orbit is given as:

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

So:

13.6 + E_n = 12.1

- \frac{13.6}{n^2} = 12.1 - 13.6 \Rightarrow - \frac{13.6}{n^2} = -1.5

Solving this, we get:

\frac{13.6}{n^2} = 1.5

Rearranging:

n^2 = \frac{13.6}{1.5} \Rightarrow n^2 = 9.0667 \Rightarrow n \approx 3

This means the electron is excited to the third energy level (n = 3).

When it transitions back to the ground state, it can do so by emitting photons corresponding to allowed transitions. The possible transitions are:

From n = 3 to n = 2
From n = 3 to n = 1
From n = 2 to n = 1

Each transition will result in a spectral line being emitted. Thus, three spectral lines will be observed.

Therefore, the correct answer is indeed three. The other options (two, four, or one) do not provide the correct number of transitions for this energy change.

Answer 2