Problem 33 - Chemistry The Central Science In Si Units

Question

Molybdenum metal must absorb radiation with an energy higher than \(7.22 \times 10^{-19} \mathrm{~J}\left({ }^{\text {" }}\right.\) energy threshold") before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(240 \mathrm{nm}\), what is the maximum possible velocity of the emitted electrons?

Step-by-Step Solution

Answer

(a) \(1.09 \times 10^{15} \text{ Hz}\); (b) \(275 \text{ nm}\); (c) \(4.83 \times 10^{5} \text{ m/s}\).

Step 1: Determine frequency threshold

The energy of a photon is given by the equation \( E = h u \), where \( E \) is the energy, \( h \) is Planck’s constant \( (6.626 \times 10^{-34} \text{ J s}) \), and \( u \) is the frequency. We'll solve for \( u \) by rearranging the equation: \( u = \frac{E}{h} \). Using \( E = 7.22 \times 10^{-19} \text{ J} \), we have \( u = \frac{7.22 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} = 1.09 \times 10^{15} \text{ Hz} \).

Step 2: Calculate wavelength for photon energy

The wavelength \( \lambda \) and frequency \( u \) of a wave are related by the speed of light equation \( c = \lambda u \). To find the wavelength, we solve for \( \lambda \): \( \lambda = \frac{c}{u} \) where \( c = 3.00 \times 10^{8} \text{ m/s} \). Using the frequency from Step 1, \( \lambda = \frac{3.00 \times 10^{8} \text{ m/s}}{1.09 \times 10^{15} \text{ Hz}} = 2.75 \times 10^{-7} \text{ m} = 275 \text{ nm} \).

Step 3: Calculate maximum velocity of emitted electrons

When light of wavelength \( \lambda = 240 \text{ nm} \) strikes the molybdenum, we first calculate the energy of this photon using \( E = \frac{hc}{\lambda} \). Convert \( \lambda \) to meters: \( 240 \text{ nm} = 240 \times 10^{-9} \text{ m} \). Now, \( E = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.00 \times 10^{8} \text{ m/s}}{240 \times 10^{-9} \text{ m}} \approx 8.29 \times 10^{-19} \text{ J} \). The kinetic energy \( KE \) of the emitted electrons is the difference between this photon energy and the threshold energy: \( KE = E_{photon} - E_{threshold} = 8.29 \times 10^{-19} \text{ J} - 7.22 \times 10^{-19} \text{ J} = 1.07 \times 10^{-19} \text{ J} \). The maximum velocity \( v \) is then found using \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \text{ kg}) \). Solving for \( v \), \( v = \sqrt{\frac{2 \times 1.07 \times 10^{-19} \text{ J}}{9.11 \times 10^{-31} \text{ kg}}} \approx 4.83 \times 10^{5} \text{ m/s} \).

Key Concepts

Photon Energy

When discussing the photoelectric effect, it all starts with the concept of photon energy. A photon is a particle of light, and its energy is what influences the emission of electrons from the surface of a metal. The energy of a single photon is determined by its frequency, given by the formula \( E = h u \), where \( E \) is the energy, \( h \) is Planck’s constant \((6.626 \times 10^{-34} \text{ Js})\), and \( u \) is the frequency of the light. This relationship shows that the energy is directly proportional to the frequency. Therefore, higher frequency light means higher energy photons. In our exercise, the molybdenum metal requires photons with an energy greater than \( 7.22 \times 10^{-19} \text{ J} \) to eject electrons. This energy value is crucial as it sets the minimum requirement for photon energy needed to cause the photoelectric effect in molybdenum.

Frequency Threshold

The frequency threshold is a pivotal point in understanding the photoelectric effect. It is the minimum frequency needed for a photon to possess enough energy to eject an electron from the surface of a metal. Using the formula \( E = h u \), we can solve for \( u \), yielding \( u = \frac{E}{h} \). In this scenario, with a threshold energy \( E \) of \( 7.22 \times 10^{-19} \text{ J} \), the frequency threshold is calculated as \( u = \frac{7.22 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} \), which equates to \( 1.09 \times 10^{15} \text{ Hz} \). This means only light with a frequency equal to or greater than this value can cause electron emission, thus illustrating the importance of frequency in the photoelectric phenomenon.

Wavelength Calculation

Wavelength is another fundamental concept in the photoelectric effect, linking light's physical properties to its energy capabilities. The wavelength \( \lambda \) is connected to frequency \( u \) through the speed of light \( c \), described by the equation \( c = \lambda u \), which can be rearranged to find wavelength: \( \lambda = \frac{c}{u} \).

• Using the threshold frequency \( u = 1.09 \times 10^{15} \text{ Hz} \), the wavelength \( \lambda \) is calculated as \( \lambda = \frac{3.00 \times 10^{8} \text{ m/s}}{1.09 \times 10^{15} \text{ Hz}} \).
• This results in \( 2.75 \times 10^{-7} \text{ m} \) or \( 275 \text{ nm} \), representing the shortest wavelength that can still induce the photoelectric effect for molybdenum.

Understanding the wavelength allows us to comprehend which part of the electromagnetic spectrum is necessary to activate the photoelectric effect on specific materials.

Maximum Electron Velocity

Once a photon successfully transfers its energy to an electron, causing it to be ejected, the maximum velocity of the electron is determined by how much energy it receives. This calculation involves finding the kinetic energy \( KE \) using the photon energy \( E_{\text{photon}} \), subtracting the threshold energy \( E_{\text{threshold}} \). Given a light with a wavelength of \( 240 \text{ nm} \, (240 \times 10^{-9} \text{ m}) \), the photon energy is calculated: \[ E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \text{ Js} \times 3.00 \times 10^{8} \text{ m/s}}{240 \times 10^{-9} \text{ m}} = 8.29 \times 10^{-19} \text{ J} \].

• The kinetic energy of the electron is \( KE = 8.29 \times 10^{-19} \text{ J} - 7.22 \times 10^{-19} \text{ J} = 1.07 \times 10^{-19} \text{ J} \).
• Using the equation \( KE = \frac{1}{2}mv^2 \), with \( m \) as the electron mass \( (9.11 \times 10^{-31} \text{ kg}) \), we find the velocity: \( v = \sqrt{ \frac{2 \times 1.07 \times 10^{-19} \text{ J}}{9.11 \times 10^{-31} \text{ kg}} } = 4.83 \times 10^{5} \text{ m/s} \).

This velocity demonstrates the potential impact of photon energy on electron movement, providing fascinating insights into the dynamic processes at the atomic level.