If change in angular momentum when an electron makes a transition corresponding to the third line of the Balmer series in Li 2+ ion is X × (h/2 π), then value of X will be

Question

If change in angular momentum when an electron makes a transition corresponding to the third line of the Balmer series in Li 2+ ion is X × (h/2 π), then value of X will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c099a75f97c0a4c7d67e02151bff09cd-.png" /> Third line of Balmer series is 5 arrow 2 Therefore, change in angular momentum is =(5 h/2 π)-(2 h/2 π)=(3 h/2 π)

Q. If change in angular momentum when an electron makes a transition corresponding to the third line of the Balmer series in $L i^{2 +}$ ion is $X \times \frac{h}{2 π}$ , then value of $X$ will be

Third line of Balmer series is $5 \to 2$
Therefore, change in angular momentum is
$= \frac{5 h}{2 π} - \frac{2 h}{2 π} = \frac{3 h}{2 π}$

Q. If change in angular momentum when an electron makes a transition corresponding to the third line of the Balmer series in Li2+ ion is X×2πh​, then value of X will be

Third line of Balmer series is 5→2 Therefore, change in angular momentum is =2π5h​−2π2h​=2π3h​

Q. If change in angular momentum when an electron makes a transition corresponding to the third line of the Balmer series in $L i^{2 +}$ ion is $X \times \frac{h}{2 π}$ , then value of $X$ will be

Third line of Balmer series is $5 \to 2$
Therefore, change in angular momentum is
$= \frac{5 h}{2 π} - \frac{2 h}{2 π} = \frac{3 h}{2 π}$