Question

Balmer series gives an equation for wavelength of visible radiation of H-spectrum as lembda=kn^2/n^2-4. The value of k in terms of R is

Answer 1

Answer:

The value of k in terms of R is $\dfrac{4}{R}$

Explanation:

Given that,

$\lambda=\dfrac{kn^2}{n^2-4}$...(I)

The light radiated in the balmer series is

When the electron jumps from upper state to lower state.

So, the Rydberg's formula is given by

$\nu=\dfrac{1}{\lambda}=(\dfrac{1}{2^2}-\dfrac{1}{n^2})$

$\nu=R(\dfrac{n^2-4}{4n^2})$

Where, R = Rydberg constant

$\lambda=\dfrac{1}{R}(\dfrac{4n^2}{n^2-4})$....(II)

From equation (I) and (II)

$\dfrac{kn^2}{n^2-4}=\dfrac{1}{R}(\dfrac{4n^2}{n^2-4})$

$k = \dfrac{4}{R}$

Hence, The value of k in terms of R is $\dfrac{4}{R}$