Question

The ratio of change in angular momentum
when an electron makes a transition
corresponding to the 3rd line of Balmer series in
Li2+ ion to that in He+ ion is​

Answer 1

To find:

Ratio of change in angular momentum when an electron makes a transition corresponding to the 3rd line of Balmer series in Li2+ ion to that in He+ ion.

Calculation:

Angular momentum of an electron at any orbit is given as :

$\therefore \: L = \dfrac{nh}{2\pi}$

So, change in Momentum ;

$= > \:\Delta L = \dfrac{ \{\Delta(n) \}h}{2\pi}$

3rd Line of Blamer series refers to transition from n = 5 to n = 2;

So , required ratio :

$\therefore \: L_{{Li}^{ + 3} } : L_{{He}^{ + 2} } = \dfrac{(5 - 2)h}{2\pi} : \dfrac{(5 - 2)h}{2\pi}$

$= > \: L_{{Li}^{ + 3} } : L_{{He}^{ + 2} } = \dfrac{(3)h}{2\pi} : \dfrac{(3)h}{2\pi}$

$= > \: L_{{Li}^{ + 3} } : L_{{He}^{ + 2} } = 1 :1$

So, final answer is:

$\boxed{ \: L_{{Li}^{ + 3} } : L_{{He}^{ + 2} } = 1 :1}$