Chapter 7, Problem 7.64QP

Interpretation Introduction

Interpretation: The transitions from higher energy states to the $n=2$ level in ${\text{He}}^{\text{+}}$ ever produce visible light is to be stated. The value of $n_2$ at which visible light is produced is to be stated.

Concept introduction: When an atom or a molecule makes electronic transition from a high energy state to a lower energy state, the spectrum of different frequencies of electromagnetic radiations is obtained known as emission spectrum.

To determine: The transitions from higher energy states to the $n=2$ level in ${\text{He}}^{\text{+}}$ ever produce visible light or not. The value of $n_2$ at which visible light is produced.

Answer to Problem 7.64QP

Solution

The transitions from higher energy states to the $n=2$ level in ${\text{He}}^{\text{+}}$ will not produce visible light.

Explanation of Solution

Explanation

Given

The transition occurs from higher energy state to $n=2$ level.

The energy of the photons emitted by one electron atoms and ions is given by the equation,

$E=\left(2.18\times {10}^{-18}\text{J}\right)Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ (1)

Where,

• $E$ is the energy of photons.
• $Z$ is an atomic number.
• $n_1$ is the lower energy state.
• $n_2$ is the higher energy state.

The values of $n_1$ and $n_2$ are considered to be 2 and $\infty$ respectively.

Substitute the values of $Z$, $n_1$ and $n_2$ in the above equation to calculate the energy of photons.

$\begin{array}{c}E=\left(2.18\times {10}^{-18}\text{J}\right){\left(2\right)}^2\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(\infty \right)}^2}\right)\\ =\left(2.18\times {10}^{-18}\text{J}\right)\times 4\times \left(\frac{1}{4}-\frac{1}{\infty }\right)\\ =8.72\times {10}^{-18}\times \left(0.25-0\right)\\ =2.18\times {10}^{-18}\text{J}\end{array}$

Hence, energy of the photons is $2.18\times {10}^{-18}\text{J}$.

The wavelength of the photons is calculated by using the equation,

$\lambda =\frac{hc}{E}$ (2)

Where,

• $\lambda$ is the wavelength.
• $h$ is the Planck’s constant $\left(6.62\times {10}^{-34}\text{J}\text{.s}\right)$.
• $c$ is the speed of light $\left(3\times {10}^8\text{m/s}\right)$.
• $E$ is the energy.

Substitute the values of $E$, $h$ and $c$ in the above equation to calculate the wavelength of photons.

$\begin{array}{c}\lambda =\frac{\left(6.62\times {10}^{-34}\text{J}\text{.s}\right)\left(3\times {10}^8\text{m/s}\right)}{2.18\times {10}^{-18}\text{J}}\\ =\frac{1.986\times {10}^{-25}\text{J}\text{.m}}{2.18\times {10}^{-18}\text{J}}\\ =9.110\times {10}^{-8}\text{m}\end{array}$

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore, the conversion of $9.110\times {10}^{-8}\text{m}$ to nm is done as,

$\begin{array}{c}9.110\times {10}^{-8}\text{m}=9.110\times {10}^{-8}\times {10}^9\text{nm}\\ =91.10\text{nm}\end{array}$

In the continuous spectrum of electromagnetic radiation, the wavelength of visible region ranges from $390\text{nm}\text{to}750\text{nm}$. Here, the wavelength of photons is $91.10\text{nm}$ that will occur in the ultra-violet region.

Hence, transitions from higher energy states to the $n=2$ level in ${\text{He}}^{\text{+}}$ will not produce visible light.

The transition in the visible light is produced when the value of wavelength must range from $390\text{nm}\text{to}750\text{nm}$. The calculated value of wavelength is $91.10\text{nm}$, hence to increase the value of wavelength the value of energy of electromagnetic radiation is to be decreases. As, it is known that,

$E\infty \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For lower value of $E$, the value of $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ must be low. As the increase in value of $n_2$, increases the value of $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$. Hence, the value of $n_2$ must be low to have wavelength of visible region. Therefore, the values of $n_1$ and $n_2$ are considered to be 2 and 3 respectively. The energy of photon is calculated by using the equation (1).

Substitute the values of $Z$, $n_1$ and $n_2$ in equation (1) to calculate the energy of photons.

$\begin{array}{c}E=\left(2.18\times {10}^{-18}\text{J}\right){\left(2\right)}^2\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)\\ =\left(2.18\times {10}^{-18}\text{J}\right)\times 4\times \left(\frac{1}{4}-\frac{1}{9}\right)\\ =8.72\times {10}^{-18}\times \left(0.25-0.11\right)\\ =8.72\times {10}^{-18}\times 0.14\text{J}\end{array}$

Simplify the above equation,

$E=1.220\times {10}^{-18}\text{J}$

The wavelength of photons emitted in the transition from $n=3\text{to}n=2$ is calculated by using the equation (2).

Substitute the values of $E$, $h$ and $c$ in the above equation to calculate the wavelength of photons.

$\begin{array}{c}\lambda =\frac{\left(6.62\times {10}^{-34}\text{J}\text{.s}\right)\left(3\times {10}^8\text{m/s}\right)}{1.22\times {10}^{-18}\text{J}}\\ =\frac{1.986\times {10}^{-25}\text{J}\text{.m}}{1.22\times {10}^{-18}\text{J}}\\ =16.27\times {10}^{-8}\text{m}\end{array}$

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore, the conversion of $16.27\times {10}^{-8}\text{m}$ to nm is done as,

$\begin{array}{c}16.27\times {10}^{-8}\text{m}=16.27\times {10}^{-8}\times {10}^9\text{nm}\\ =162.7\text{nm}\end{array}$

The wavelength of photons is still lesser than the wavelength required producing visible light. Hence, transitions from higher energy states to the $n=2$ level in ${\text{He}}^{\text{+}}$ will never produce visible light.

Conclusion

Transitions from higher energy states to the $n=2$ level in ${\text{He}}^{\text{+}}$ will never produce visible light as the calculated wavelength is lesser than the wavelength required producing visible light.