Problem 1 - College Physics

Question

Response of the eye. The human eye is most sensitive to green light of wavelength \(505 \mathrm{nm} .\) Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina. (a) What is the frequency of this photon? (b) How much energy (in joules and eV) does it deliver to the receptor cells? (c) To appreciate what a small amount of energy this is, calculate how fast a typical bacterium of mass \(9.5 \times 10^{-12} \mathrm{~g}\) would move if it had that much energy.

Step-by-Step Solution

Answer

(a) Frequency is approximately \(5.94 \times 10^{14} \mathrm{Hz}\). (b) Energy is \(3.94 \times 10^{-19} \mathrm{J}\) or \(2.46 \mathrm{eV}\). (c) Bacterium moves at \(9.1 \times 10^{-3} \mathrm{m/s}\).

Step 1: Convert Wavelength to Frequency

First, we use the formula that relates the speed of light \(c\), frequency \(f\), and wavelength \(\lambda\): \(c = f \cdot \lambda\). Given \(\lambda = 505\, \mathrm{nm} = 505 \times 10^{-9} \, \mathrm{m}\) and \(c = 3 \times 10^{8} \, \mathrm{m/s}\), we solve for \(f\): \[f = \frac{c}{\lambda} = \frac{3 \times 10^8}{505 \times 10^{-9}} \, \mathrm{Hz}\,\]\\[f \approx 5.94 \times 10^{14} \, \mathrm{Hz}\].

Step 2: Calculate Photon Energy in Joules

The energy \(E\) of a photon is given by \(E = h \cdot f\), where \(h = 6.626 \times 10^{-34} \, \mathrm{J \cdot s}\). Using \(f = 5.94 \times 10^{14} \, \mathrm{Hz}\), we find:\[E = 6.626 \times 10^{-34} \times 5.94 \times 10^{14} \, \mathrm{J}\,\]\\[E \approx 3.94 \times 10^{-19} \, \mathrm{J}\].

Step 3: Convert Energy from Joules to Electron Volts

We can convert energy in joules to electron volts (eV) using the conversion factor \(1 \, \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J}\). Thus,\[E = \frac{3.94 \times 10^{-19}}{1.602 \times 10^{-19}} \, \mathrm{eV}\,\]\\[E \approx 2.46 \, \mathrm{eV}\].

Step 4: Calculate Bacterium Speed with Photon Energy

The kinetic energy \(K\) of an object is given by \(K = \frac{1}{2}mv^2\), where \(m = 9.5 \times 10^{-12} \, \mathrm{g} = 9.5 \times 10^{-15} \, \mathrm{kg}\), and \(v\) is speed. Setting \(K = 3.94 \times 10^{-19} \, \mathrm{J}\), we solve for \(v\):\[3.94 \times 10^{-19} = \frac{1}{2}(9.5 \times 10^{-15})v^2\,\]\[v^2 = \frac{2 \times 3.94 \times 10^{-19}}{9.5 \times 10^{-15}}\,\]\[v \approx \sqrt{8.3 \times 10^{-5}}\,\]\[v \approx 9.1 \times 10^{-3} \, \mathrm{m/s}\].

Key Concepts

Wavelength and Frequency

Light, like many other forms of radiation, travels in waves. The distance between consecutive crests of a wave is known as the wavelength, often measured in nanometers (nm). In our exercise, we're interested in green light, which has a wavelength of \(505 \, \text{nm}\).
The frequency, on the other hand, refers to how many wave cycles occur in a second, measured in Hertz (Hz). The frequency and wavelength are related by the speed of light \(c\), using the formula:

• \(c = f \cdot \lambda\)

Here, \(c\) is the speed of light \(3 \times 10^8 \, \text{m/s}\), \(f\) is the frequency, and \(\lambda\) is the wavelength in meters. Converting \(505 \, \text{nm}\) to meters gives us \(505 \times 10^{-9} \, \text{m}\), allowing us to solve for the frequency:

• \(f = \frac{3 \times 10^8}{505 \times 10^{-9}}\)
• \(f \approx 5.94 \times 10^{14} \, \text{Hz}\)

Understanding this relationship helps us decipher how often light waves impact our eyes and consequently how we perceive different colors.

Photon Energy

A photon is a packet of light's energy, and understanding its energy is crucial in physics. The energy \(E\) of a photon can be calculated using its frequency \(f\) and Planck's constant \(h\):

• \(E = h \cdot f\)

Here, \(h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\), a constant that relates energy and frequency of a photon.
Substituting the frequency from before, \(f = 5.94 \times 10^{14} \, \text{Hz}\), we calculate:

• \(E = 6.626 \times 10^{-34} \times 5.94 \times 10^{14} \, \text{J}\)
• \(E \approx 3.94 \times 10^{-19} \, \text{J}\)

This incredibly tiny amount of energy highlights the efficiency of our eyes, as a single photon can trigger a response in the retina's cells. Moreover, energy can also be expressed in electronvolts (eV), with one eV equivalent to \(1.602 \times 10^{-19} \, \text{J}\). Converting our photon energy, we have:

• \(E \approx \frac{3.94 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.46 \, \text{eV}\)

Visual Perception

Human vision is a fascinating and complex process, heavily reliant on how our eyes interpret light. The retina, located at the back of the eye, is key to this mechanism. It consists of two primary types of photosensitive cells: rods and cones.

• Rods: Highly sensitive to low levels of light and enable night vision. They are predominantly responsive to green light (505 nm).
• Cones: Responsible for color vision and operate well in bright light.

In extremely dim environments, such as the dark room scenario in the exercise, our rods are crucial. They can be stimulated by a single photon of green light, despite its low energy, enabling our brain to "see" even in low-light conditions. This ability demonstrates our evolutionary adaptation and reliance on both types of photoreceptor cells to not only perceive color but also to survive in various lighting conditions.

Kinetic Energy Calculations

Kinetic energy (KE) is the energy an object possesses due to its motion. If you know the energy, you can determine how fast the object would move. The relationship is expressed by the formula:

• \(K = \frac{1}{2}mv^2\)

where \(m\) is the mass and \(v\) is the velocity.
In our exercise, we are asked to determine how fast a bacterium could move if it absorbed the energy of a photon \(E = 3.94 \times 10^{-19} \, \text{J}\). Given the mass \(m = 9.5 \times 10^{-12} \, \text{g} = 9.5 \times 10^{-15} \, \text{kg}\), solving for \(v\) involves:

• Set \(K = 3.94 \times 10^{-19} \, \text{J}\)
• Use \(3.94 \times 10^{-19} = \frac{1}{2}(9.5 \times 10^{-15})v^2\)
• Find \(v\): \(v \approx \sqrt{8.3 \times 10^{-5}} \approx 9.1 \times 10^{-3} \, \text{m/s}\)

This calculation emphasizes how minuscule amounts of energy (such as that from a photon) can translate into measurable physical effects, especially in microorganisms or small particles.