Question

In a certain electronic transition in the hydrogen atom from an initial state 1 to final state 2 the difference in the orbital radius is 24 times the first bohr radius

Answer 1

Completing Question :
Find the possible initial and final states?


Final Answer : i=5 , f = 7
i= 1. , f = 5
Solution :
Steps and Understanding :

1) We know that,
radius of an Hydrogen orbit in a state( n)is proportional to n^2 .

2) Hence,
R(n) = kn^2
where k is constant
=>
$r(n) = k {n}^{2}$

3) Now,
Let initial state be (i)
Final State be (f).

According to question,
$r(f) - r( i) = 24 \times r(1)\\ k \times {f}^{2} - k \times {i}^{2} = k \times {1}^{2} \\ = > {f}^{2} - {i}^{2} = 24$
4) Now,
This is Diophantine Equation,
Case : 1
${f}^{2} - {i}^{2} = 24 \\ = > (f - i) \times (f + i) = 2 \times 12$

$f - i = 2 - - - - (1) \\ f + i = 12 - - - - (2)$
On solving, we get
f = 7 ,i = 5

5) Case : 2

${f}^{2} - {i}^{2} = 24 \\ = > (f - i) \times (f + i) = 4 \times 6 \\ = > f - i = 4 - - - (3) \\ f + i = 6 - - - (4)$

On Solving (3) and (4), we get
f = 5 ,i = 1 ,

6) Other Cases will not give integer solutions to (i, f) .

So, Finally Possible initial and Final states are

Initial State= 5 , Final State = 7
Initial State = 1 , Final State = 5 .

Answer 2

Explanation:

but both options are given