Question

calculate the frequency of the photon of limiting line of loomer series

fast answer I will mark as brilliant ​

Answer 1

Solution:

Limiting line of Balmer's series refers to the transition from  n₁ = ∞ to n₂ = 2.

We know that:

$\tt{\dfrac{1}{\lambda} = R \Big( \dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\Big)}\\ \\ \\ \implies \tt{\dfrac{1}{\lambda}= R\Big( \dfrac{1}{2^2} - \dfrac{1}{\infty^2}\Big)}\\ \\ \\ \implies \tt{\dfrac{1}{\lambda}= R\Big(\dfrac{1}{4} \Big)} \\ \\ \\ \implies \tt{\lambda = \dfrac{4}R}\\ \\ \\ \tt{We\ know\ that:\ \ \ f = \dfrac{c}{\lambda}} \\ \\ \\ \implies \tt{f = \dfrac{cR}{4} = \dfrac{3\times 10^8 \times 1096780}{4} = 8.23 \times 10^{14}\ \ Hz}\\ \\ \\ \boxed{\tt{f = 8.23 \times 10^{14}\ \ Hz}}$

Hope this helps : )

Answer 2

Hope it helps you.........