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

The minimum energy needed to eject an electron from molybdenum is $7.22\times {10}^{-19}J$. The sufficient wavelength of radiation needed to provide a photon of this minimum energy is 275 nm. The maximum possible kinetic energy of the emitted electrons when molybdenum is irradiated with light of wavelength 120 nm is $9.34\times {10}^{-19}J$.

Step-by-step solution

(a) Find the minimum energy needed to eject an electron

First, we need to find the minimum energy (E_min) required to eject an electron from molybdenum by using the given frequency and the energy equation. Given frequency (f) = $1.09\times {10}^{15}{s}^{-1}$ Planck's constant (h) = $6.626\times {10}^{-34}Js$ Now, use the energy equation E = hf: E_min = hf = $6.626\times {10}^{-34}Js\times 1.09\times {10}^{15}{s}^{-1}$ E_min = $7.22\times {10}^{-19}J$ The minimum energy needed to eject an electron from molybdenum is $7.22\times {10}^{-19}J$.

(b) Determine the wavelength needed to provide the minimum energy

To find the wavelength (λ_min) that will provide the minimum energy, use the de Broglie wavelength equation: λ_min = c/f Given speed of light (c) = $3\times {10}^{8}m/s$ Now, use de Broglie's wavelength equation: λ_min = $3\times {10}^{8}m/s$ / $1.09\times {10}^{15}{s}^{-1}$ λ_min = $2.75\times {10}^{-7}m$ or $275nm$ So the sufficient wavelength of radiation needed to provide a photon of the minimum energy required is 275 nm.

(c) Calculate the maximum kinetic energy of emitted electrons

Given the wavelength of radiation (λ) = $120nm=1.2\times {10}^{-7}m$, now we need to find the energy of the incident light photon (E_incident) and then the maximum kinetic energy of the emitted electrons using the work-energy principle. First, find the frequency (f_incident): f_incident = c/λ = $3\times {10}^{8}m/s$ / $1.2\times {10}^{-7}m$ f_incident = $2.5\times {10}^{15}{s}^{-1}$ Now, find the energy of the incident photon (E_incident): E_incident = hf_incident = $6.626\times {10}^{-34}Js$ * $2.5\times {10}^{15}{s}^{-1}$ E_incident = $1.6565\times {10}^{-18}J$ Now, using the work-energy principle, calculate the maximum kinetic energy of the emitted electrons (K_max): K_max = E_incident - E_min = $1.6565\times {10}^{-18}J$ - $7.22\times {10}^{-19}J$ K_max = $9.34\times {10}^{-19}J$ The maximum possible kinetic energy of the emitted electrons when molybdenum is irradiated with light of wavelength 120 nm is $9.34\times {10}^{-19}J$.

Key concepts

Planck's constant

Planck's constant, denoted as 'h', is a fundamental constant in quantum mechanics that plays a pivotal role in the explanation of phenomena such as the photoelectric effect. It is the proportionality constant connecting the energy (E) of a photon to its frequency (f), through the well-known relation,

$E=hf$.

This equation implies that the energy of a photon is directly proportional to its frequency, with Planck's constant as the factor of proportionality. The value of Planck's constant is approximately $6.626\times {10}^{-34}J\cdot s$, which is exceedingly small, signifying the quantum nature of energy at the atomic and subatomic levels.

In the context of the photoelectric effect, Planck's constant enables us to calculate the energy of photons needed to eject electrons from the surface of a material. Without this constant, the precise quantification of energy based on frequency would not be possible, thus hindering our understanding of the interaction between light and matter.

De Broglie wavelength

The de Broglie wavelength concept is a fundamental principle in quantum mechanics, introducing wave-particle duality. It posits that particles such as electrons can exhibit wave-like properties, including wavelength and frequency. The de Broglie wavelength ($\lambda$) of a particle is given by the equation

$\lambda =\frac{h}{p}$,

where 'h' is Planck's constant and 'p' is the momentum of the particle. The momentum can be expressed as the product of the mass (m) and velocity (v) of the particle, thus $p=mv$.

For photons, which are massless particles of light, the momentum is directly related to the frequency (or inversely to the wavelength) of the photon. To find the wavelength that corresponds to a particular energy, you can manipulate de Broglie's equation in conjunction with the energy relation $E=hf$, which gives us the wavelength of the radiation needed for effects like photoemission. This relation is critical for determining the wavelengths that can induce the photoelectric effect, as seen in the provided solution for the minimum energy required to eject an electron.

Kinetic energy of electrons

The kinetic energy (KE) of an electron is the energy it possesses due to its motion, and it can be calculated using the equation

$KE=\frac{1}{2}m{v}^2$,

where 'm' and 'v' represent the mass and velocity of the electron, respectively. In the context of the photoelectric effect, electrons are emitted from a material surface when struck by photons of sufficient energy.

The maximum kinetic energy of an emitted electron is the energy of the incident photon minus the minimum energy required to eject the electron from the material, which is known as the work function ($\varphi$). The energy conservation principle can then be expressed as

$K_{\text{max}}=E_{\text{incident}}-\varphi$.

The exercise provided calculates the $K_{\text{max}}$ by subtracting the minimum energy needed to eject an electron ($E_{\text{min}}$) from the energy of the incident photon ($E_{\text{incident}}$). This is crucial for understanding how varying the wavelength and frequency of incident radiation affects the kinetic energy of the emitted electrons and the overall photoelectric effect.