Electrons in a certain energy leven n=n1, can emit 3 spectral lines. When they are in another energy level, n=n2. They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio of
A
4:3
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B
3:4
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C
2:1
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D
1:2
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Solution
The correct option is C4:3 Number of emission spectral lines, N=n(n−1)2 ∴3=n1(n1−1)2, in first case.
n21−n1−6=0 or (n1−3)(n1+2)=0 Take positive root. ∴n1=3
Again, 6=n2(n2−1)2, in second case. n22−n2−12=0 or (n2−4)(n2+3)=0. Take positive root, or n2=4