Question

Maximum emitted wavelength by photon during a transition in H-atom that produce a visible light

Answer 1

Answer: The maximum emitted wavelength by photon of H-atom will be 656nm.

Explanation: The visible light comes under the Balmer Series of H-atom having n = 2.

We are asked to calculate the maximum wavelength, which states that the energy should be minimum because,

$E=\frac{hc}{\lambda}$

here, $E\propto \frac{1}{\lambda}$

Since, energy and wavelength follow inverse relationship, the minimum energy corresponds to the maximum wavelength.

Minimum energy is calculated when the emission is taking place from the next highest energy level to lower energy level which corresponds to n = 3.

Using Rydberg's Equation,

$\frac{1}{\lambda}=R\left ( \frac{1}{n_f^2}-\frac{1}{n_i^2}\right )$

where, R = Rydberg's Constant ($1.097\times10^7m^{-1}$)

$n_f$ = Final energy level, here it is 2

$n_i$ = Initial energy level, here it is 3

$\frac{1}{\lambda}=1.097\times10^7m^{-1} \left ( \frac{1}{2^2}-\frac{1}{3^2}\right )$

$\lambda=6.565\times10^{-7}m$

Conversion factor: $1nm=10^{-9}m$

$\lambda=656nm$