Question

Molybdenum metal must absorb radiation with an energy higher than $7.22\times {10}^{-19}\mathrm{J}$ ( "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?

Short answer

The frequency threshold for emission of electrons is $1.09\times {10}^{15}\mathrm{Hz}$. The wavelength of radiation that provides a photon of this energy is $275\mathrm{nm}$. The maximum possible velocity of the emitted electrons when molybdenum is irradiated with light of wavelength $240\mathrm{nm}$ is $4.38\times {10}^5\mathrm{m}/\mathrm{s}$.

Step-by-step solution

a) Frequency threshold for emission of electrons#

Using the energy of a photon equation, we can solve for the frequency threshold: $E_{threshold}=h{\nu }_{threshold}$ Solving for ${\nu }_{threshold}$: ${\nu }_{threshold}=\frac{E_{threshold}}{h}$ Given $E_{threshold}=7.22\times {10}^{-19}J$ and $h=6.63\times {10}^{-34}Js$, we can calculate the frequency threshold: ${\nu }_{threshold}=\frac{7.22\times {10}^{-19}}{6.63\times {10}^{-34}}=1.09\times {10}^{15}\mathrm{Hz}$ The frequency threshold for emission of electrons is $1.09\times {10}^{15}\mathrm{Hz}$.

b) Wavelength of radiation for the given photon energy#

Using the wavelength and frequency relationship, we can solve for the wavelength of radiation that provides the given photon energy: $c=\lambda \nu$ Solving for $\lambda$: $\lambda =\frac{c}{\nu }$ Given $c=3\times {10}^8\mathrm{m}/\mathrm{s}$ and ${\nu }_{threshold}=1.09\times {10}^{15}\mathrm{Hz}$, we can calculate the wavelength of radiation: $\lambda =\frac{3\times {10}^8}{1.09\times {10}^{15}}=2.75\times {10}^{-7}\mathrm{m}=275\mathrm{nm}$ The wavelength of radiation that provides a photon of this energy is $275\mathrm{nm}$.

c) Maximum possible velocity of emitted electrons#

Given the wavelength of irradiated light, we can first calculate the energy of a photon at $240\mathrm{nm}$: $E_{photon}=h\nu =\frac{hc}{\lambda }$ Using $h=6.63\times {10}^{-34}Js$, $c=3\times {10}^8\mathrm{m}/\mathrm{s}$ and $\lambda =240\times {10}^{-9}\mathrm{m}$, we can determine the energy of a photon: $E_{photon}=\frac{(6.63\times {10}^{-34})(3\times {10}^8)}{240\times {10}^{-9}}=8.29\times {10}^{-19}J$ Now we can calculate the kinetic energy of emitted electrons: $K{E}_{max}=E_{photon}-E_{threshold}$ Using $E_{photon}=8.29\times {10}^{-19}J$ and $E_{threshold}=7.22\times {10}^{-19}J$: $K{E}_{max}=8.29\times {10}^{-19}-7.22\times {10}^{-19}=1.07\times {10}^{-19}J$ Finally, we can find the maximum possible velocity of emitted electrons using the kinetic energy equation: $K{E}_{max}=\frac{1}{2}m{v}^2$ Solving for $v_{max}$: $v_{max}=\sqrt{\frac{2K{E}_{max}}{m}}$ Assuming the mass of an electron is $m=9.11\times {10}^{-31}\mathrm{kg}$: $v_{max}=\sqrt{\frac{2(1.07\times {10}^{-19})}{9.11\times {10}^{-31}}}=4.38\times {10}^5\mathrm{m}/\mathrm{s}$ The maximum possible velocity of the emitted electrons is $4.38\times {10}^5\mathrm{m}/\mathrm{s}$.

Key concepts

Frequency Threshold

Imagine light as tiny packets of energy known as photons. For an electron to be ejected from a metal surface, each photon must carry enough energy to overcome a certain barrier. This is where the concept of the "frequency threshold" comes into play. The frequency threshold is the minimum frequency that incoming photons must have to trigger the ejection of electrons from a material's surface.

The energy of a photon is directly related to its frequency through the equation:

• $E=hu$, where $E$ is the energy, $h$ is Planck’s constant $6.63\times {10}^{-34}\text{Js}$, and $u$ is the frequency.
• Thus, the threshold frequency $u_{threshold}$ is the frequency at which the photon's energy equals the energy threshold necessary to eject an electron.
• For molybdenum, this was calculated as $1.09\times {10}^{15}\text{Hz}$.

Without meeting this frequency, no electron emission occurs, regardless of the light's intensity. Simply put, the frequency must hit or exceed this threshold for electrons to be liberated.

Wavelength of Radiation

Another way to understand photon energy is through its wavelength. Photons of light can be described in terms of their wavelength, which relates to both their frequency and energy.

The wavelength is inversely proportional to frequency:

• $c=\lambda u$, where $c$ is the speed of light $3\times {10}^8\text{m/s}$, $\lambda$ is the wavelength, and $u$ is the frequency.
• Thus, as the frequency increases (higher energy), the wavelength becomes shorter.
• For molybdenum, the wavelength associated with its energy threshold was calculated to be $275\text{nm}$.

This shows that specific photon energies correlate to specific wavelengths. In general, shorter wavelengths carry more energy and are more likely to eject electrons.

Kinetic Energy of Electrons

When light shines on a metal and meets the threshold frequency, electrons are ejected. These electrons gain kinetic energy as a result of a process known as the photoelectric effect. The concept of *kinetic energy* here is about how fast these electrons can move after being ejected.

The maximum kinetic energy of an emitted electron is given by:

• $K{E}_{max}=E_{photon}-E_{threshold}$
• Where $K{E}_{max}$ is the maximum kinetic energy, $E_{photon}$ is the energy of the incoming photon, and $E_{threshold}$ is the energy threshold for the material.
• For molybdenum, when light with a wavelength of $240\text{nm}$ was used, the kinetic energy of electrons was found to be $1.07\times {10}^{-19}\text{J}$.
• The velocity of these electrons was calculated using: $v_{max}=\sqrt{\frac{2K{E}_{max}}{m}}$, where $m$ is the mass of an electron.
• This resulted in a velocity of $4.38\times {10}^5\text{m/s}$.

Electron velocity and kinetic energy are directly influenced by the energy provided by the photons. Hence, higher photon energy increases both the energy and velocity of emitted electrons.