Question

In a certain electronic transition in the hydrogen atom from an initial state i to final state f the difference in the orbit radius is seven times the first bohr orbit

Answer 1

Question Completion :Find Initial and Final States.(i, f)
Final Answer : Initial State = 3
Final State = 4
$\bf {SOLUTION}$
Steps and Understanding :
1)We know that radius of an orbit for a particular atom is proportional to n^2 where n is the orbit number.
Then,
R = k n^2
where k is constant for a particular atom.

2) According to the question,

R(f) - R(i) =7* R(1)
=> k (f)^2 - k(i)^2 = 7*k*(1)^2
$= > {f}^{2} - {i}^{2} = 7 \\ = > (f - i)(f + i) = 1 \times 7$
Since,7 is prime number and f> i,
so,
$f - i = 1 \: \: \: \: - - - (1) \\ (f + i) = 7 - - - - - (2)$

On solving above equations,
$f = 4 \: \: \: and \: i \: = 3$

3) So, Required initial state n = 3
Final State n = 4

Answer 2

See your answer in the attachment my friend
Hope this will help you.