krit.club logo

Nuclear and Quantum Physics - Discrete Energy and Photoelectric Effect

Grade 12IBPhysics

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

🔑Concepts

•

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

•

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.

•

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.

•

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.

•

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

•

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

•

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.

•

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

•

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=1.989×10−254.5×10−7≈4.42×10−19 JE = \frac{1.989 \times 10^{-25}}{4.5 \times 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.