Problem 2 - University Physics with Modern 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 electron volts) 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) 5.94 × 10¹⁴ Hz, (b) 3.94 × 10⁻¹⁹ J or 2.46 eV, (c) 9.11 × 10⁻³ m/s.

Step 1: Convert Wavelength to Frequency

We start by using the relationship between wavelength and frequency. The speed of light is given by the equation: \[ c = \lambda \cdot f \] where \( c \) is the speed of light (\( 3 \times 10^8 \ \mathrm{m/s} \)), \( \lambda \) is the wavelength (505 nm or \( 505 \times 10^{-9} \ \mathrm{m} \)), and \( f \) is the frequency. Solving for frequency, we have: \[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{505 \times 10^{-9}} \approx 5.94 \times 10^{14} \ \mathrm{Hz} \].

Step 2: Calculate Energy in Joules

The energy of a photon can be calculated using the equation: \[ E = h \cdot f \] where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \ \mathrm{Js} \)). Using the frequency from the previous step, \( f = 5.94 \times 10^{14} \ \mathrm{Hz} \), we find: \[ E = 6.626 \times 10^{-34} \cdot 5.94 \times 10^{14} \approx 3.94 \times 10^{-19} \ \mathrm{J} \].

Step 3: Convert Energy to Electron Volts

To convert the energy from joules to electron volts, we use the conversion factor \( 1 \ \mathrm{eV} = 1.602 \times 10^{-19} \ \mathrm{J} \). Therefore, the energy in electron volts is: \[ \frac{3.94 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.46 \ \mathrm{eV} \].

Step 4: Calculate Velocity of a Bacterium

The kinetic energy of a bacterium can be described by the equation: \[ K.E. = \frac{1}{2} m v^2 \] where \( m \) is the mass of the bacterium (\( 9.5 \times 10^{-12} \ \mathrm{g} \) or \( 9.5 \times 10^{-15} \ \mathrm{kg} \)), and \( v \) is the velocity. We equate this energy to the photon's energy: \[ \frac{1}{2} \cdot 9.5 \times 10^{-15} \cdot v^2 = 3.94 \times 10^{-19} \]. Solving for \( v \), we get: \[ v^2 = \frac{2 \cdot 3.94 \times 10^{-19}}{9.5 \times 10^{-15}} \approx 8.29 \times 10^{-5} \]. Taking the square root gives: \[ v \approx 9.11 \times 10^{-3} \ \mathrm{m/s} \].

Key Concepts

Wavelength and Frequency

Understanding the relationship between wavelength and frequency is essential when studying light properties, such as those emitted by a photon. The speed of light (\(c\), approximately \(3 \times 10^8\) \(\mathrm{m/s}\)), remains constant. However, both wavelength (\(\lambda\)) and frequency (\(f\)) determine its characteristics. This relationship is expressed by the equation: \[ c = \lambda \cdot f \] Wavelength is the distance between successive peaks of a wave and is usually measured in meters. Frequency is the number of waves that pass a given point per second, measured in hertz (\(\mathrm{Hz}\)). Green light with a wavelength of 505 \(\mathrm{nm}\) can be converted into frequency using the formula: \[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{505 \times 10^{-9}} \approx 5.94 \times 10^{14} \ \mathrm{Hz} \] Knowing how to switch between wavelength and frequency is crucial when analyzing photon energy, which we will explore further in conjunction with Planck's Constant.

Planck's Constant

Planck's constant (\(h\)) is a fundamental constant used in quantifying energy at the quantum level. It connects the energy of a photon to its frequency. Originating from Max Planck's work on black body radiation, its value is approximately \(6.626 \times 10^{-34} \ \mathrm{Js}\). Employing Planck's constant, the energy (\(E\)) of a photon can be calculated as follows: \[ E = h \cdot f \] Utilizing the frequency determined earlier, \(f = 5.94 \times 10^{14} \ \mathrm{Hz}\), we can find the energy of a green photon: \[ E = 6.626 \times 10^{-34} \cdot 5.94 \times 10^{14} \approx 3.94 \times 10^{-19} \ \mathrm{J} \] Energy is often expressed in joules, but when dealing with atomic scales, electron volts (\(\mathrm{eV}\)) might be more intuitive. Conversion to electron volts involves dividing by the conversion factor \(1 \ \mathrm{eV} = 1.602 \times 10^{-19} \ \mathrm{J}\), yielding: \[ \frac{3.94 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.46 \ \mathrm{eV} \] Planck's constant is pivotal in understanding quantum mechanics, underpinning the explanation of photon interactions like those that trigger receptor cells in your eyes.

Kinetic Energy

Kinetic energy (\(K.E.\)) is the energy possessed by an object due to its motion. For a mass \(m\) with velocity \(v\), it is mathematically depicted by the formula: \[ K.E. = \frac{1}{2} m v^2 \] To appreciate the extremely small energy carried by a photon, compare it to something tangible. Imagine transferring that energy to a bacterium of mass \(9.5 \times 10^{-15} \ \mathrm{kg}\). By equating the photon's energy to the bacterium's kinetic energy and solving for velocity, we find: \[ \frac{1}{2} \cdot 9.5 \times 10^{-15} \cdot v^2 = 3.94 \times 10^{-19} \] Thus: \[ v^2 = \frac{2 \cdot 3.94 \times 10^{-19}}{9.5 \times 10^{-15}} \approx 8.29 \times 10^{-5} \] Taking the square root, \(v \approx 9.11 \times 10^{-3} \ \mathrm{m/s}\). This illustrates that even a minuscule photon energy could move an ultralight object like a bacterium at observable speeds, demonstrating the photon's capacity to impart motion through its kinetic energy.