Question

A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to the longest wavelength amongst them?

Answer 1

Wavelength = 975 Å = 975 × 10⁻¹⁰ m.

From the Ground state to the excited state, this means, n₁ = 1

∴ 1/λ = R(1/n₁² - 1/n₂²)

∴ 1/975 × 10⁻¹⁰ = R(1/1 - 1/n²)

10¹⁰ = 975R(n² - 1)/n²

10¹⁰n² = 975Rn² - 975R

10¹⁰n² = 1.09 × 10⁷(975n² - 975)

1000n² = 1.09 × 975n² - 975 × 1.09

1000n² = 1062.75n² - 1062.75

1062.75n² - 1000n² = 1062.75

62.75n² = 1062.75

n² = 16.94

∴ n = 4.12

∴ n ≈ 4

This means, electrons jumps from the ground state to the 4th level.

Now, Using the Formula,

No. of Spectral lines = (n₂ - n₁ + 1)(n₂ - n₁)/2

= (4 - 0 + 1)(4 - 0)/2

= 5 × 4/2

= 5 × 2

= 10

Therefore, no. of spectral lines is 10.

Longest Transition is from n=1  to n = 4.

Hope it helps.