Question

Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Ȧ. Number of spectral lines in the resulting spectrum emitted will be:
1. 2
2. 6
3. 10
4. 3

Answer 1

Correct Answer - Option 2 : 6

Ans: Option 2)

CONCEPT:

• Bohr's third postulate: An electron radiates energy in the form of radiation when an electron jumps from an orbit of higher energy to an orbit of lower energy.
  • The energy of a radiated photon is equal to the difference of energies of electrons of two states.
• In a normal hydrogen atom, the electron revolves in the first orbit for which the principal quantum number is 1.

EXPLANATION:

We know that energy of photon is given as,

\(E= \frac{hc}{λ }\)

where E is the energy of the photon, λ is the wavelength, h is Planck's constant, c is the speed of light.

Here, λ = 975 A = 975× 10^-10 m

E = \(\frac{6.67\times 10^{-34}\times 3\times 10^{8}}{975\times 10^{^{-10}}}\) 

E = 12.75eV

Number of spectral line is given as

N_E = \(\frac{n(n-1)}{2}\)

To find n we use Bohr's formula for wavenumber and wavelength of spectral line of hydrogen,

\(\frac{1}{\lambda }=R\left ( \frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}} \right )\)

where R is Rydberg's constant.

here n₁=1

so by using above formula we get 

n₂ = n = 4

 ∴ Number of spectral line N_E = \(\frac{n(n-1)}{2}\)= \(\frac{4\left ( 4-1 \right )}{2}\)

N_E = 6

So the correct answer is option 2)