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An electron in one Bohr orbit with principal quantum number n = n1 can produce 3 distinct spectral lines on de-excitation. In another orbit with principal quantum number n = n2, it can produce 6 distinct spectral lines. What is the ratio of the orbital speeds of the electron in these two orbits?
- 4: 3
- 3: 4
- 2: 1
- 1: 2
Correct answer: 4: 3
Solution
Number of spectral lines = n(n-1)/2, so 3 lines -> n1 = 3 and 6 lines -> n2 = 4. Orbital speed v is proportional to 1/n, so v1:v2 = (1/3):(1/4) = 4:3.
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