Question

An electron jumps from 4th excited state to ground state in hydrogen atom, then the number of lines obtained in UV region will be

Answer 1

Total no of line is n(n+1)/2
=>n = n4-n1 =4-1=3
No of line=3(3+1)/2=6

Answer 2

An electron jumps from 4th excited state to ground state in hydrogen atom, then the number of lines obtained in UV region will be 6 spectral lines.

Explanation:

• Number of a spectral line in an atom is given by the formula,
• No spectral lines=$\frac{(n1-n2)(n1-n2+1)}{2}$
• n2-the orbit where the electron enters.
• n2-the orbit where from where the electron comes out.
• So, according to the question, n2=1and n1=4.
• Then, a number of spectral lines when an electron jumps from 4th orbit to first orbit is,$= \frac{4-1)(4-1+1)}{2}$ =6 spectral lines.
• Every jump is associated with two states - the initial one and the final one. In case of transition from 4 to 1, there are 4 states available. The number of ways in which you can pick 2 states would be 4C2 or C(4,2) = 6.

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