Question

Molybdenum metal must absorb radiation with a minimum frequency of $1.09\times {10}^{15}{\mathrm{s}}^{-1}$ before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of $120\mathrm{nm}$, what is the maximum possible kinetic energy of the emitted electrons?

Short answer

(a) The minimum energy needed to eject an electron is $7.23\times {10}^{-19}\mathrm{J}$. (b) The wavelength of radiation that provides this energy is 276 nm. (c) The maximum possible kinetic energy of the emitted electrons is $9.38\times {10}^{-19}\mathrm{J}$.

Step-by-step solution

Find the minimum energy needed to eject an electron

To find the minimum energy needed to eject an electron, we can use the given frequency and the equation $E=h\nu$. We are given the frequency $\nu =1.09\times {10}^{15}{\mathrm{s}}^{-1}$ and the Planck's constant $h=6.63\times {10}^{-34}\mathrm{Js}$. Now, we can calculate the energy: $E=h\nu =(6.63\times {10}^{-34}\mathrm{Js})(1.09\times {10}^{15}{\mathrm{s}}^{-1})=7.23\times {10}^{-19}\mathrm{J}$ So, the minimum energy needed to eject an electron is $7.23\times {10}^{-19}\mathrm{J}$.

Find the wavelength of radiation that provides this energy

To find the wavelength of radiation, we'll use the formula $\lambda =\frac{c}{\nu }$, where $c$ is the speed of light and $\nu$ is the given frequency. $c=3.00\times {10}^8\mathrm{m}/\mathrm{s}$ Now, we can find the wavelength: $\lambda =\frac{c}{\nu }=\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.09\times {10}^{15}{\mathrm{s}}^{-1}}=2.76\times {10}^{-7}\mathrm{m}$ To convert this to nanometers, multiply by ${10}^9$: $\lambda =2.76\times {10}^{-7}\mathrm{m}\times {10}^9\mathrm{nm}/\mathrm{m}=276\mathrm{nm}$ So, the wavelength of radiation that provides this energy is 276 nm.

Find the maximum possible kinetic energy of the emitted electrons

To find the maximum kinetic energy of the emitted electrons, we'll use the Einstein's photoelectric equation: $K_{max}=h\nu -\varphi$ We already know the value of $h\nu$. Now, we need to find the work function $\varphi$. We know that the threshold wavelength ${\lambda }_{threshold}$ is 276 nm. The work function can be related to this threshold wavelength using the same equations as before: $\varphi =h{\nu }_{threshold}=\frac{hc}{{\lambda }_{threshold}}=\frac{(6.63\times {10}^{-34}\mathrm{Js})(3.00\times {10}^8\mathrm{m}/\mathrm{s})}{2.76\times {10}^{-7}\mathrm{m}}=7.23\times {10}^{-19}\mathrm{J}$ Now, we are given that the actual wavelength of the irradiated light is 120 nm. So, we need to find the energy of the incident photon at this wavelength: $E_{incident}=\frac{hc}{\lambda }=\frac{(6.63\times {10}^{-34}\mathrm{Js})(3.00\times {10}^8\mathrm{m}/\mathrm{s})}{120\times {10}^{-9}\mathrm{m}}=1.66\times {10}^{-18}\mathrm{J}$ Using the Einstein's photoelectric equation, we can find the maximum kinetic energy of the emitted electrons: $K_{max}=E_{incident}-\varphi =1.66\times {10}^{-18}\mathrm{J}-7.23\times {10}^{-19}\mathrm{J}=9.38\times {10}^{-19}\mathrm{J}$ So, the maximum possible kinetic energy of the emitted electrons is $9.38\times {10}^{-19}\mathrm{J}$.

Key concepts

Molybdenum

Molybdenum is a metal with intriguing properties, notably its role in the photoelectric effect. The photoelectric effect is the emission of electrons from a metal when it absorbs light. For molybdenum, this requires light of a particular frequency. The frequency needed is at least $1.09\times {10}^{15}{\text{s}}^{-1}$, which means molybdenum interacts with high-frequency electromagnetic radiation. Metals like molybdenum are crucial when studying electron dynamics because they have well-defined work functions—the minimum energy required to eject an electron—in the photoelectric effect mechanism.
Understanding this property of molybdenum helps us explore how electrons are bound within an atom and how they can be released when light is introduced.

Energy of Photon

The energy of a photon is calculated using the formula $E=hu$, where $h$ is Planck's constant ($6.63\times {10}^{-34}\text{Js}$), and $u$ is the frequency of the light. Photons are the basic units of light, and they carry energy depending on their frequency. For the given frequency of $1.09\times {10}^{15}{\text{s}}^{-1}$, the energy of a photon is $7.23\times {10}^{-19}\text{J}$. This energy is vital to understand because it's the minimum energy necessary to start releasing electrons from molybdenum via the photoelectric effect.

• This photon energy is equivalent to the work function of molybdenum.
• Energy calculations like these are foundational in quantum mechanics since they relate quantum principles to observable phenomena.

Electron Ejection

Electron ejection occurs when a metal absorbs sufficient energy that surpasses a threshold, leading to electron release. The threshold energy for molybdenum is $7.23\times {10}^{-19}\text{J}$.Using the formula $E_{photon}=hu$, molybdenum requires photons with this minimum energy to overcome its work function—the inherent energy barrier keeping electrons bound to the metal's surface.
When the absorbing frequency of the light matches this threshold, electrons emit, leading to the photoelectric effect where energy conversion from light to kinetic electron energy occurs.To further explore the concept:

• Only photons equal to or exceeding this energy can eject electrons.
• This phenomenon illustrates how energy transfers in quantum systems manifest as physical actions like electron movement.

Kinetic Energy of Electrons

The maximum kinetic energy ($K_{max}$) of electrons emitted from a metal is a function of the photon energy minus the work function, expressed in Einstein's photoelectric equation: $K_{max}=hu-\varphi$. For molybdenum, consider the incident wavelength of 120 nm to solve for $K_{max}$.By substituting the wavelength into the formula $E=\frac{hc}{\lambda }$, we find $E_{incident}=1.66\times {10}^{-18}\text{J}$.

• The calculation for maximum kinetic energy then becomes $9.38\times {10}^{-19}\text{J}$.
• This value indicates how excess energy translates into the movement of electrons, after breaking free from the metal surface.

Understanding kinetic energy in this context sheds light on fundamental quantum physics principles, driving technological applications such as photovoltaic cells and photoelectron spectroscopy.