Chapter 5, Problem 111A

Interpretation Introduction

Interpretation:

The energy of hydrogen atom is $-6.05\times {10}^{-20}\text{J}$ when the electron is present in the $n=6$ orbit and $-2.18\times {10}^{-18}\text{J}$ when the electron is present in the $n=1$ orbit. Determine the wavelength of photon emitted when the electron drops from $n=6$ to $n=1$ orbit. Use the following values: $h=6.626\times {10}^{-34}\text{J}\text{.s}$ and $c=3.00\times {10}^8\text{m/s}$.

Concept introduction:

The wavelength of a light is the distance between the corresponding points in two adjacent light cycles while frequency of light is a number of cycles of the light that pass a given point in one second.

Formula used:

$\text{E}=\text{h}v\text{and}v=\frac{c}{\text{λ}}$

where, E represents the energy of the light, h represents the Plank’s constant, c is the speed of the light, →is the wavelength of light and →is the frequency of light.

Answer to Problem 111A

The wavelength of a photon emitted is $9.38\times {10}^{-8}\text{m}$.

Explanation of Solution

To determine the wavelength of photon emitted, difference the first energy level and the sixth energy level and determine the energy change. The speed of light is $c=3.00\times {10}^8\text{m/s}$, Plank’s constant is $h=6.626\times {10}^{-34}\text{J}\text{.s}$, energy at the first level $n=1$ is $-2.18\times {10}^{-18}\text{J}$ and energy at the sixth level $n=6$ is $-6.05\times {10}^{-20}\text{J}$.

Substitute the values in the equation given as:

$\begin{array}{c}\text{λ}=\frac{\text{hc}}{\left({\text{E}}_1-{\text{E}}_2\right)}\\ =\frac{\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\times \left(3.00\times {10}^8\text{m/s}\right)}{\left(-2.18\times {10}^{-18}\text{J}\right)-\left(-6.05\times {10}^{-20}\text{J}\right)}\\ =9.38\times {10}^{-8}\text{m}\end{array}$

Hence, the wavelength of a photon emitted is $9.38\times {10}^{-8}\text{m}$.

Conclusion

The wavelength of a photon emitted is $9.38\times {10}^{-8}\text{m}$.