This chapter covers the key formulas for the photoelectric effect, the photon picture of light, and the de Broglie wavelength of matter. You’ll find the relations linking energy, momentum, work function, stopping potential, threshold frequency, and the wave nature of particles.
Each formula is grouped by topic, with symbol meanings, when-to-use guidance, and original worked examples. For detailed explanations and derivations, visit the Class 12 Physics notes page.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Maximum kinetic energy of photoelectrons | \( K_{\max} = h\nu – \phi_0 \) |
| Stopping potential from maximum KE | \( K_{\max} = e V_0 \) |
| Einstein’s photoelectric equation (stopping potential form) | \( e V_0 = h\nu – \phi_0 \) |
| Threshold frequency (from work function) | \( \nu_0 = \dfrac{\phi_0}{h} \) |
| Stopping potential vs. frequency (linear relation) | \( V_0 = \dfrac{h}{e}\nu – \dfrac{\phi_0}{e} \) |
| Energy of a photon | \( E = h\nu = \dfrac{hc}{\lambda} \) |
| Momentum of a photon | \( p = \dfrac{h\nu}{c} = \dfrac{h}{\lambda} \) |
| de Broglie wavelength of a particle | \( \lambda = \dfrac{h}{p} = \dfrac{h}{mv} \) |
| Electron volt to joule conversion | \( 1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J} \) |
All Formulas, Grouped by Topic
Photoelectric Effect
Maximum kinetic energy of photoelectrons:
\[ K_{\max} = h\nu – \phi_0 \]
where \( h \) is Planck’s constant, \( \nu \) the frequency of incident light, and \( \phi_0 \) the work function of the metal. This is Einstein’s photoelectric equation (NCERT, p. 8).
Relation between stopping potential and maximum KE:
\[ K_{\max} = e V_0 \]
The stopping potential \( V_0 \) is the minimum retarding potential that reduces the photocurrent to zero. (NCERT, p. 5) Einstein’s photoelectric equation in terms of stopping potential:
\[ e V_0 = h\nu – \phi_0 \quad (\nu \ge \nu_0) \]
Threshold frequency:
\[ \nu_0 = \frac{\phi_0}{h} \]
Below \( \nu_0 \) no photoelectrons are emitted, no matter how intense the light. (NCERT, p. 9) Stopping potential vs. frequency (linear relation):
\[ V_0 = \frac{h}{e}\nu – \frac{\phi_0}{e} \]
The slope of the \( V_0 \) vs. \( \nu \) graph is \( h/e \), independent of the metal. (NCERT, p. 9)
Photon Picture of Light
Energy of a photon:
\[ E = h\nu = \frac{hc}{\lambda} \]
All photons of a given frequency (or wavelength) carry the same energy. (NCERT, p. 10) Momentum of a photon:
\[ p = \frac{h\nu}{c} = \frac{h}{\lambda} \]
Photons have momentum even though they are massless. (NCERT, p. 10)
de Broglie Wavelength (Wave Nature of Matter)
de Broglie wavelength of a moving particle:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where \( m \) is the mass and \( v \) the speed of the particle. (NCERT, p. 11)
Unit Conversion
Electron volt to joule:
\[ 1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J} \]
What Each Symbol Means
| Symbol | What it means | SI unit |
|---|---|---|
| \( h \) | Planck’s constant | J s |
| \( \nu \) | Frequency of incident radiation | Hz (s⁻¹) |
| \( \lambda \) | Wavelength of radiation / de Broglie wavelength | m |
| \( c \) | Speed of light in vacuum | m/s |
| \( \phi_0 \) | Work function of the metal | J (or eV) |
| \( K_{\max} \) | Maximum kinetic energy of an emitted photoelectron | J (or eV) |
| \( e \) | Elementary charge (magnitude of electron charge) | C |
| \( V_0 \) | Stopping potential | V |
| \( \nu_0 \) | Threshold frequency | Hz |
| \( m \) | Mass of a particle | kg |
| \( v \) | Speed of a particle | m/s |
| \( p \) | Momentum | kg m/s |
| \( E \) | Energy of a photon | J (or eV) |
When to Use Each Formula
| Formula / Situation | When to use | Condition |
|---|---|---|
| \( K_{\max} = h\nu – \phi_0 \) | To find the maximum kinetic energy of photoelectrons when the frequency of incident light is known. | \( \nu \ge \nu_0 \) |
| \( e V_0 = h\nu – \phi_0 \) | To relate stopping potential to frequency; also used to find work function or Planck’s constant from experimental data. | \( \nu \ge \nu_0 \) |
| \( \nu_0 = \phi_0 / h \) | To find the threshold frequency from the work function, or vice versa. | Always valid |
| \( V_0 = (h/e)\nu – \phi_0/e \) | To interpret the graph of stopping potential vs. frequency; slope gives \( h/e \), intercept gives \( -\phi_0/e \). | \( \nu \ge \nu_0 \) |
| \( E = h\nu = hc/\lambda \) | To find the energy of a photon given its frequency or wavelength. | Always valid for photon |
| \( p = h\nu/c = h/\lambda \) | To find the momentum of a photon. | Always valid for photon |
| \( \lambda = h/p = h/(mv) \) | To find the de Broglie wavelength of a moving particle (electron, proton, etc.). | Non-relativistic speeds preferred; for relativistic use \( p = \gamma m v \) |
Worked Examples
Example 1: Maximum kinetic energy of photoelectrons
Light of frequency \( 8.0 \times 10^{14}\) Hz falls on a metal whose work function is \( 2.50\) eV. Find the maximum kinetic energy of the emitted photoelectrons in joules and in eV.
Step 1: Convert work function to joules if needed.
\[ \phi_0 = 2.50\ \text{eV} = 2.50 \times 1.602 \times 10^{-19}\ \text{J} = 4.005 \times 10^{-19}\ \text{J} \]
Step 2: Photon energy \( E = h\nu \).
\[ E = (6.626 \times 10^{-34}\ \text{J s})(8.0 \times 10^{14}\ \text{Hz}) = 5.301 \times 10^{-19}\ \text{J} \]
Step 3: Apply Einstein’s equation.
\[ K_{\max} = E – \phi_0 = 5.301 \times 10^{-19}\ \text{J} – 4.005 \times 10^{-19}\ \text{J} = 1.296 \times 10^{-19}\ \text{J} \]
Step 4: Convert back to eV.
\[ K_{\max} = \frac{1.296 \times 10^{-19}\ \text{J}}{1.602 \times 10^{-19}\ \text{J/eV}} = 0.809\ \text{eV} \]
Final answer: \( K_{\max} = 1.30 \times 10^{-19}\ \text{J} \) (or 0.809 eV).
Example 2: Finding the stopping potential
A metal surface has a work function of \( 1.90\) eV. When illuminated with light of wavelength \( 450\) nm, photoelectrons are emitted. Find the stopping potential.
Step 1: Convert wavelength to frequency (or use \( E = hc/\lambda \)).
\[ \nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8\ \text{m/s}}{450 \times 10^{-9}\ \text{m}} = 6.667 \times 10^{14}\ \text{Hz} \]
Step 2: Photon energy in eV.
\[ E = h\nu = (6.626 \times 10^{-34})(6.667 \times 10^{14}) = 4.417 \times 10^{-19}\ \text{J} \]
\[ E = \frac{4.417 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.757\ \text{eV} \]
Step 3: Maximum KE = \( E – \phi_0 = 2.757 – 1.90 = 0.857\) eV.
Step 4: Stopping potential \( V_0 = K_{\max} / e = 0.857\) V (since \( K_{\max} \) in eV equals numerical value of \( V_0 \) in volts).
Final answer: \( V_0 = 0.857\) V.
Example 3: de Broglie wavelength of an electron
An electron is accelerated through a potential difference of 100 V. Find its de Broglie wavelength. (Assume non-relativistic; electron mass \( m_e = 9.11 \times 10^{-31}\) kg, \( e = 1.602 \times 10^{-19}\) C.)
Step 1: Kinetic energy gained = \( eV = 100\) eV = \( 100 \times 1.602 \times 10^{-19} = 1.602 \times 10^{-17}\) J.
Step 2: Speed from kinetic energy: \( \frac{1}{2} m v^2 = K \).
\[ v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1.602 \times 10^{-17}}{9.11 \times 10^{-31}}} = \sqrt{3.517 \times 10^{13}} = 5.931 \times 10^6\ \text{m/s} \]
Step 3: Momentum \( p = mv = 9.11 \times 10^{-31} \times 5.931 \times 10^6 = 5.403 \times 10^{-24}\) kg m/s.
Step 4: de Broglie wavelength.
\[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.403 \times 10^{-24}} = 1.226 \times 10^{-10}\ \text{m} = 0.1226\ \text{nm} \]
Final answer: \( \lambda = 0.123\) nm (comparable to X-ray wavelengths).
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Forgetting that \( K_{\max} \) must be non-negative; using \( K_{\max} = h\nu – \phi_0 \) even when \( \nu \lt \nu_0 \) | Photoelectric emission only occurs if \( h\nu \ge \phi_0 \). If \( \nu \lt \nu_0 \), \( K_{\max} \) is meaningless; no emission. | Check that the calculated \( K_{\max} \) is positive. If negative, the frequency is below threshold. |
| Confusing work function \( \phi_0 \) with threshold frequency \( \nu_0 \) | \( \phi_0 \) is energy (J or eV); \( \nu_0 = \phi_0/h \) is frequency (Hz). They are different quantities. | Always verify units: work function is energy, threshold frequency is 1/time. |
| Using the wrong sign in Einstein’s equation: writing \( h\nu + \phi_0 \) instead of \( h\nu – \phi_0 \) | Energy conservation: photon energy = work function + maximum kinetic energy. So \( K_{\max} = h\nu – \phi_0 \). | If your \( K_{\max} \) comes out larger than the photon energy, the sign is wrong. |
| Forgetting to convert eV to J when using SI units in formulas | All constants (like \( h \)) are in SI units. If work function is given in eV, convert to J before using \( h \) in J s. | Check that units cancel: \( h \) (J s) × \( \nu \) (Hz) gives J. If \( \phi_0 \) is in eV, convert to J or keep consistent. |
| Applying the de Broglie formula \( \lambda = h/(mv) \) to photons | For photons, \( p = h/\lambda \), so \( \lambda = h/p \) holds, but \( m \) is zero; use \( p = h\nu/c \). | If you get a finite mass for a photon, you’ve used the wrong momentum expression. |
Frequently Asked Questions
What is the stopping potential and how is it related to the maximum kinetic energy?
The stopping potential \( V_0 \) is the minimum negative potential applied to the collector that stops the most energetic photoelectrons. The relation is \( K_{\max} = e V_0 \). So if you measure \( V_0 \), you directly know \( K_{\max} \) in eV (numerically equal).
Why does the photoelectric effect prove the particle nature of light?
The effect shows that light energy is absorbed in discrete packets (photons) each of energy \( h\nu \). Classical wave theory predicted that the kinetic energy of electrons should increase with intensity, and that there should be no threshold frequency. Einstein’s photon model explains all observations: the maximum kinetic energy depends only on frequency, not intensity, and emission is instantaneous.
What is the de Broglie wavelength of an electron accelerated through 1 V?
For an electron accelerated through \( V \) volts, the de Broglie wavelength is given by \( \lambda = h/\sqrt{2meV} \). For \( V = 1\) V, \( \lambda \approx 1.226\) nm. This is a useful shortcut: \( \lambda (\text{nm}) = 1.226 / \sqrt{V} \).
How do I convert between eV and Joules?
Use the conversion \( 1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J} \). To convert energy from eV to J, multiply by \( 1.602 \times 10^{-19} \). To convert J to eV, divide by the same number.
Reference: NCERT Class 12 Physics textbook, chapter Dual Nature of Radiation and Matter.
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