Problem 82 - Chemistry The Central Science

Question

A photocell is a device used to measure the intensity of light. In a certain experiment, when light of wavelength \(630 \mathrm{nm}\) is directed onto the photocell, electrons are emitted at the rate of \(2.6 \times 10^{-12} \mathrm{C} / \mathrm{s}\) (coulombs per second). Assume that each photon that impinges on the photocell emits one electron. How many photons per second are striking the photocell? How much energy per second is the photocell absorbing?

Step-by-Step Solution

Answer

The number of photons striking the photocell per second is approximately \(1.625 \times 10^{7} \: photons/s\), and the energy absorbed by the photocell per second is approximately \(5.12 \times 10^{-12} \: J/s\).

Step 1: Identify the givens

We are given the following information: - Wavelength of the light: \(λ = 630 \: nm\) - Rate of electrons emitted: \(2.6 \times 10^{-12} \: C/s\) We need to find two things: 1. The number of photons striking the photocell per second. 2. The energy per second the photocell is absorbing.

Step 2: Convert wavelength to meters

We need to convert the given wavelength from nanometers to meters. \[λ = 630 \: nm \times \frac{1 \: m}{10^9 \: nm} = 630 \times 10^{-9} \: m\]

Step 3: Find the rate of photons striking the photocell

We know that each photon impinging on the photocell emits one electron. As the charge of an electron is known, we can calculate the number of electrons emitted per second and equate it to the number of photons per second. Charge of an electron: \(e = 1.6 \times 10^{-19} \: C\) Rate of photons striking the photocell per second: \[\frac{2.6 \times 10^{-12} \: C/s}{1.6 \times 10^{-19} \: C/electron} = \frac{2.6}{1.6} \times 10^{7} electrons/s\] The number of photons striking the photocell per second is approximately \(1.625 \times 10^{7} \: photons/s\).

Step4: Find the energy of a single photon

Now, we can calculate the energy of a single photon using the Planck's constant (\(h = 6.626 \times 10^{-34} \: Js\)) and the speed of light (\(c = 3 \times 10^8 \: m/s\)). The energy of a single photon is given by: \[E = \frac{hc}{λ}\] We substitute the values we found in steps 2 and 3 and calculate the energy of a single photon: \[E = \frac{6.626 \times10^{-34} \: Js \times 3 \times 10^8 \: m/s}{630 \times 10^{-9} \: m} = 3.15 \times 10^{-19} \: J\]

Step 5: Find the energy per second the photocell is absorbing

Finally, to find the energy per second the photocell is absorbing, we multiply the energy of a single photon by the number of photons per second: Energy per second: \(3.15 \times 10^{-19} \: J/photon \times 1.625 \times 10^{7} \: photons/s = 5.12 \times 10^{-12} \: J/s\) The energy absorbed by the photocell per second is approximately \(5.12 \times 10^{-12} \: J/s\).

Key Concepts

Wavelength Conversion

Understanding wavelength conversion is crucial when dealing with light and its interactions with materials, such as in a photocell. The wavelength of light is usually given in nanometers (nm) but can be converted to meters (m) for use in formulas in physics. This conversion is straightforward and is done by multiplying the value in nanometers by the conversion factor of \(10^{-9}\), as one meter is equivalent to \(10^9\) nanometers.

For instance, in the provided exercise, we take the wavelength of \(630 \text{nm}\) and convert it to meters, resulting in \(630 \times 10^{-9} \text{m}\). With the correct units, we're able to apply the relevant formulas for calculations in the photoelectric effect.

Photoelectric Effect

The photoelectric effect is a phenomenon in which electrons are emitted from a material's surface when it is exposed to electromagnetic radiation of a certain frequency. This concept is pivotal in understanding how photocells work. When light with sufficient energy strikes the photocell, it causes electrons to be emitted at a rate that depends on the intensity of the light.

In our exercise, the rate of electron emission is tied to the light's wavelength and intensity, allowing us to determine the number of photons hitting the photocell per second. By knowing the charge of each electron, we can calculate the concurrent rate of photons, linking the macroscopic observation of current to the microscopic events involving individual photons and electrons. The core mechanism explained by the photoelectric effect not only applies to photocells but is also essential in understanding various technologies including solar panels and photo detectors.

Planck's Constant

Planck's constant (\(h\)) is a fundamental constant in quantum mechanics. This constant is crucial for calculations related to the photoelectric effect and other quantum phenomena. It represents a measure of the quantization of energy, showing that energy is transferred in discrete units, or quanta, rather than continuously. The value of Planck's constant is approximately \(6.626 \times 10^{-34} \text{Js}\).

When dealing with photocells, Planck's constant is used to calculate the energy associated with a single photon based on its frequency or wavelength. As shown in our exercise, the energy of a photon (\(E\)) is given by the formula \(E = \frac{hc}{\lambda}\), where \(c\) is the speed of light and \(\lambda\) is the wavelength. Thus, Planck's constant directly assists in determining how much energy per second the photocell is absorbing, highlighting its importance in both theoretical and practical applications in physics.