Chapter 16, Problem 33Q

(a)

To determine

The longest wavelength (in nanometer) that can dislodge an electron from a negative hydrogen ion. Given that the amount of energy ($E$) that dislodges an electron from a negative hydrogen ion is $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$.

Answer to Problem 33Q

Solution:

$1.657\times {10}^3\text{ nm}$

Explanation of Solution

Given data:

The amount of energy ($E$) that dislodges an electron from a negative hydrogen ion is $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$.

Formula used:

Write the expression for the energy of a photon:

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

Here, $E$ is energy in Joules, $c$ is the speed of light, $h$ is Planck’s constant and $\lambda$ is the wavelength in meters.

Explanation:

As Planck’s constant, $h$, is $6.626\times {10}^{-34}\text{ Js}$ and the amount of energy ($E$) that dislodges an electron from a negative hydrogen ion is $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$.

Also consider the speed of light, $3\times {10}^8\text{ m/s}$.

Refer to the expression for the energy of a photon.

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

Substitute $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$ for E, $3\times {10}^{8}m/s$ for c, and $6.626\times {10}^{-34}\text{Js}$.

$\begin{array}{c}\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}=\frac{\left(6.626\times {10}^{-34}\text{Js}\right)\left(3\times {10}^{8}m/s\right)}{\lambda }\\ \lambda =\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{\left(\text{1}{\text{.2×10}}^{\text{-19}}\right)}\\ =1.657\times {10}^3\text{ nm}\end{array}$

Conclusion:

Hence, the longest wavelength that will emit an electron from the negative hydrogen ion will be $1.657\times {10}^3\text{ nm}$.

(b)

To determine

The part of the electromagnetic spectrum in which the wavelength calculated in part (a) will lie. Given that the amount of energy ($E$) that dislodges an electron from a negative hydrogen ion is $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$.

Answer to Problem 33Q

Solution:

This wavelength will lie in the infrared region of the electromagnetic spectrum.

Explanation of Solution

Introduction:

The electromagnetic spectrum is broadly classified into three classes, which are ultraviolet region, visible, and infrared regions. The wavelength increases while the energy decreases from the ultraviolet to infrared region. The visible region lies between 400 to 700 nm.

Explanation:

The wavelength calculated in part (a) came out to be $1.657\times {10}^3\text{ nm}$ or 1657 nm, which is way beyond 700 nm on the electromagnetic spectrum. The wavelength of light beyond 700 nm is of the infrared radiations. Thus, the wavelength will lie in the infrared region.

Conclusion:

Therefore, the wavelength that will emit an electron from the negative hydrogen ion will lie in the infrared region of the electromagnetic spectrum.

(c)

To determine

Whether a photon of visible light will be able to dislodge the same electron from a negative hydrogen ion. Given that the amount of energy ($E$) that dislodges an electron from a negative hydrogen ion is $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$.

Answer to Problem 33Q

Solution:

Yes, a photon of the visible light will be able to dislodge an extra electron.

Explanation of Solution

Introduction:

The visible light is called so because it can be seen through naked eyes. Wavelength other than from the visible region will only be seen through special filters. The visible part lies between the infrared and ultraviolet regions in the electromagnetic spectrum.

Explanation:

The energy level of the visible light is higher than that of the infrared radiations. As the minimum energy required to expel an electron lies in the infrared region, a photon of visible light will definitely be able to do so. The photon of visible light will be able to excite the electron much quicker than that of the infrared region.

Conclusion:

Therefore, a photon of the visible region will be able to excite an electron from the negative hydrogen ion much quickly than by the one lying in the infrared region.

(d)

To determine

The reason for the opacity of the photosphere that contains negative hydrogen ions for the visible light, but not for the infrared light. Given that the amount of energy ($E$) that dislodges an electron from a negative hydrogen ion is $\text{1}{\text{.2×10}}^{\text{-19}}\text{ J}$.

Answer to Problem 33Q

Solution:

The photosphere is opaque (invisible) for the visible light because it is of high energy and gets easily absorbed in order to dislodge electrons. The infrared light carries low energy and does not pass easily through the photosphere; therefore, they are not opaque.

Explanation of Solution

Introduction:

The light that gets absorbed easily through a medium becomes opaque (invisible), but it remains capable enough to excite electrons in the material it passes through. The light that fails to excite electrons from a medium cannot pass through the same.

Explanation:

The energy carried by the infrared radiations is far less than that carried by the ultraviolet and/or the visible light. Higher energy is able to excite the electrons through the photosphere, which is primarily made up of negative hydrogen ions. As the visible light excites electrons, it gets easily absorbed in the photosphere, thus becomes invisible or opaque.

The infrared light of longer wavelength (or less energy) than the one calculated in part (a) will fail to excite the electrons from the photosphere. They will not pass through the photosphere, thus will become visible or opaque to naked eyes.

Conclusion:

Therefore, the infrared light is opaque to the photosphere while visible light is not because the former is of less energy and does not pass through the photosphere while the latter passes through and excites the electrons so as to become invisible.