Question

Calculate the frequency of the light emitted when an electron in a hydrogen atom makes each of the following transitions. Express in inverse seconds.

a) n=4?n=3

b) n=5?n=1

c) n=5?n=4

d) n=6?n=5

Answer 1

The expression for energy of light emitted when an electron in a hydrogen atom makes transition from n2 to n1 level can be written as follows:

ΔE = -b (1/n₂² – 1/n₁²)

Where, b is constant having value 2.18 x 10^-18 J

Since, ΔE = hν

Where, h = plank’s constant = 6.63 x 10^-34 JS

ν = frequency

Thus, expression for frequency can be written as follows:

ν = -(b/h) (1/n₂² – 1/n₁²)

Above expression can be used to calculate frequency of light in given transitions as follows:

1. n=4 to n=3

Ans:

ν = -(b/h) (1/n₂² – 1/n₁²)

ν = -(2.18 x 10^-18 J / 6.63 x 10^-34 JS) (1/4² – 1/3²)

ν = 0.016 x 10¹⁶ S^-1

2. n=5 to n=1

Ans:

ν = -(b/h) (1/n₂² – 1/n₁²)

ν = -(2.18 x 10^-18 J / 6.63 x 10^-34 JS) (1/5² – 1/1²)

ν = 0.316 x 10¹⁶ S^-1

3. n=5 to n=4

Ans:

ν = -(b/h) (1/n₂² – 1/n₁²)

ν = -(2.18 x 10^-18 J / 6.63 x 10^-34 JS) (1/5² – 1/4²)

ν = 0.0074 x 10¹⁶ S^-1

4. n=6 to n=5

Ans:

ν = -(b/h) (1/n₂² – 1/n₁²)

ν = -(2.18 x 10^-18 J / 6.63 x 10^-34 JS) (1/6² – 1/5²)

ν = 0.00402 x 10¹⁶ S^-1