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Nuclear and Quantum Physics - Discrete Energy and Photoelectric Effect

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Discrete Energy Levels: Electrons within an atom can only occupy specific, quantized energy states. They cannot exist between these levels.

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Photons: Light and other electromagnetic radiation consist of discrete packets of energy called photons. The energy of a single photon is given by E=hfE = hf.

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Transitions: When an electron moves from a higher energy level (E2E_2) to a lower one (E1E_1), it emits a photon of energy hf=E2−E1hf = E_2 - E_1. Conversely, it absorbs a photon to move to a higher state.

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Emission Spectra: A series of bright lines on a dark background produced when excited electrons drop to lower energy states, emitting specific wavelengths of light unique to that element.

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Absorption Spectra: A continuous spectrum with dark lines where specific wavelengths have been absorbed by a gas, exciting electrons to higher states.

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Photoelectric Effect: The emission of electrons (photoelectrons) from a metal surface when light of a sufficiently high frequency is incident upon it.

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Work Function (Φ\Phi): The minimum energy required for an electron to escape from the surface of a metal. It is a property of the material.

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Threshold Frequency (f0f_0): The minimum frequency of incident radiation required to cause the photoelectric effect, where hf0=Φhf_0 = \Phi.

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Stopping Potential (VsV_s): The potential difference required to stop the most energetic photoelectrons from reaching the anode, such that Emax=eVsE_{max} = eV_s.

📐Formulae

E=hfE = hf

c=fλc = f \lambda

E=hcλE = \frac{hc}{\lambda}

ΔE=hf=∣Efinal−Einitial∣\Delta E = hf = |E_{final} - E_{initial}|

hf=Φ+Emaxhf = \Phi + E_{max}

Φ=hf0\Phi = h f_0

Emax=eVsE_{max} = e V_s

💡Examples

Problem 1:

Calculate the energy of a photon of blue light with a wavelength of 450 nm450\text{ nm}. (Use h=6.63×10−34 J sh = 6.63 \times 10^{-34}\text{ J s} and c=3.00×108 m s−1c = 3.00 \times 10^8\text{ m s}^{-1})

Solution:

Using the formula E=hcλE = \frac{hc}{\lambda}: E=(6.63×10−34 J s)×(3.00×108 m s−1)450×10−9 mE = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m s}^{-1})}{450 \times 10^{-9} \text{ m}} E=1.989×10−254.5imes10−7≈4.42×10−19 JE = \frac{1.989 \times 10^{-25}}{4.5 imes 10^{-7}} \approx 4.42 \times 10^{-19} \text{ J}

Explanation:

The energy of a photon is inversely proportional to its wavelength. By converting the wavelength to meters (1 nm=10−9 m1\text{ nm} = 10^{-9}\text{ m}), we can calculate the energy in Joules.

Problem 2:

A metal has a work function of 2.28 eV2.28\text{ eV}. If light of frequency 1.0×1015 Hz1.0 \times 10^{15}\text{ Hz} shines on the metal, what is the maximum kinetic energy of the emitted photoelectrons in eV\text{eV}? (Use h=4.14×10−15 eV sh = 4.14 \times 10^{-15} \text{ eV s})

Solution:

  1. Calculate incident photon energy in eV\text{eV}: Ephoton=hf=(4.14×10−15 eV s)×(1.0×1015 Hz)=4.14 eVE_{photon} = hf = (4.14 \times 10^{-15} \text{ eV s}) \times (1.0 \times 10^{15} \text{ Hz}) = 4.14 \text{ eV}.
  2. Apply the photoelectric equation: Emax=hf−ΦE_{max} = hf - \Phi.
  3. Emax=4.14 eV−2.28 eV=1.86 eVE_{max} = 4.14 \text{ eV} - 2.28 \text{ eV} = 1.86 \text{ eV}.

Explanation:

According to Einstein's photoelectric equation, the maximum kinetic energy is the difference between the energy supplied by the photon and the energy required to liberate the electron (work function).

Problem 3:

An electron drops from the n=3n=3 state to the n=2n=2 state in a hydrogen atom. If the energy of the n=3n=3 state is −1.51 eV-1.51\text{ eV} and the n=2n=2 state is −3.40 eV-3.40\text{ eV}, find the frequency of the emitted photon.

Solution:

  1. Find the energy difference: ΔE=E3−E2=−1.51 eV−(−3.40 eV)=1.89 eV\Delta E = E_3 - E_2 = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}.
  2. Convert energy to Joules: ΔE=1.89×1.60×10−19 J=3.024×10−19 J\Delta E = 1.89 \times 1.60 \times 10^{-19} \text{ J} = 3.024 \times 10^{-19} \text{ J}.
  3. Use f=ΔEhf = \frac{\Delta E}{h}: f=3.024×10−19 J6.63×10−34 J s≈4.56×1014 Hzf = \frac{3.024 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} \approx 4.56 \times 10^{14} \text{ Hz}.

Explanation:

When an electron transitions between levels, the emitted photon carries the exact energy difference between those states. Frequency is then determined by the Planck relation.

Discrete Energy and Photoelectric Effect Revision - Grade 11 Physics IB