Chapter 6, Problem 127AP

Interpretation Introduction

Interpretation:

The wavelength of the light emitted when hydrogen undergoes a transition from $n=236$ to $n=235$ is to be calculated and the region in which it falls is to be determined.

Concept introduction:

When an electron jumps from one energy state to another, it is termed as the transition of electron. Energy is either lost or absorbed during the transition of electron. The transition energy $\left(\Delta E\right)$ is determined as follows:

Delta $\text{E}=-\text{2}.\text{18}\times \text{1}{0}^{-\text{18}}\left({\text{Z}}^{\text{2}}/{\text{n}}^{\text{2}}\right)$ where $\text{Z}$ is the nuclear charge and is equal to the atomic number of hydrogen that is $\text{1}.00\text{8}$.

$\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Here, $n_f^{}$ is the excited state and $n_i^{}$ is the ground state.

The energy of a photon can be expressed as:

$\begin{array}{c}\text{E}=\text{h}\nu \\ \text{or, E}=\frac{\text{hc}}{\lambda }\end{array}$

Here, E is the energy of the photon, $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is frequency.