CUET UG Physics Modern Physics Photoelectric Effect Einsteins Equation Threshold Frequency

Must-Know

• Einstein’s photoelectric equation is ( K_{\text{max}} = h\nu - \phi_0 ), where ( K_{\text{max}} ) is maximum kinetic energy of emitted electrons, ( h ) is Planck’s constant (6.626 × 10⁻³⁴ J s), ( \nu ) is frequency of incident light, and ( \phi_0 ) is work function of the metal.
• Work function (( \phi_0 )) is the minimum energy required to eject an electron from the metal surface; for sodium, ( \phi_0 \approx 2.75 \, \text{eV} ) (verify from NCERT).
• Threshold frequency (( \nu_0 )) is the minimum frequency of incident radiation below which no photoelectrons are emitted, given by ( \nu_0 = \frac{\phi_0}{h} ).
• For a metal with work function 4.2 eV, threshold frequency is ( \nu_0 = \frac{4.2 \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 1.01 \times 10^{15} \, \text{Hz} ).
• Photoelectric emission is instantaneous (time lag < 10⁻⁹ s), even at very low light intensity, supporting particle nature of light.
• Maximum kinetic energy of photoelectrons depends on frequency of incident light, not intensity; doubling intensity increases photocurrent, not ( K_{\text{max}} ).
• Stopping potential (( V_0 )) is the minimum retarding potential needed to stop the fastest photoelectrons: ( eV_0 = K_{\text{max}} = h\nu - \phi_0 ).
• If incident light frequency equals threshold frequency, ( K_{\text{max}} = 0 ) and stopping potential ( V_0 = 0 ).
• The photoelectric effect supports quantum theory and contradicts classical wave theory, which predicts emission at any frequency with sufficient intensity.
• Slope of ( V_0 ) vs ( \nu ) graph is ( \frac{h}{e} ), a universal constant (same for all metals), approximately ( 4.14 \times 10^{-15} \, \text{V s} ).
• Intercept on frequency axis in ( V_0 ) vs ( \nu ) graph gives threshold frequency ( \nu_0 ); intercept on voltage axis gives ( -\frac{\phi_0}{e} ).
• Cut-off wavelength (( \lambda_0 )) is related to threshold frequency by ( \lambda_0 = \frac{c}{\nu_0} ); for ( \nu_0 = 1.0 \times 10^{15} \, \text{Hz} ), ( \lambda_0 = 300 \, \text{nm} ).
• Photons are quanta of light energy; energy of one photon is ( E = h\nu = \frac{hc}{\lambda} ), where ( c = 3 \times 10^8 \, \text{m/s} ).
• In photoelectric effect, one photon ejects one electron (photon-electron interaction is one-to-one).
• Intensity of light is proportional to number of photons per unit area per unit time; higher intensity → more photoelectrons → higher photocurrent.
• Metals like cesium and potassium have low work functions and are used in photoelectric cells due to sensitivity to visible light.
• Einstein was awarded the Nobel Prize in Physics in 1921 specifically for his explanation of the photoelectric effect, not for relativity.
• Photoelectric effect occurs only when ( \nu \geq \nu_0 ), regardless of intensity; e.g., red light (low ( \nu )) may not eject electrons even if intense, but weak UV light (high ( \nu )) can.
• The equation ( h\nu = \phi_0 + \frac{1}{2}mv_{\text{max}}^2 ) is the energy conservation form of Einstein’s photoelectric equation.
• Work function can be expressed in eV or joules: 1 eV = ( 1.6 \times 10^{-19} \, \text{J} ).

Difficulty Level

Intermediate — requires understanding of both conceptual physics and numerical applications of Einstein’s equation and graph interpretations.

Common CUET Traps

• Trap: Assuming higher intensity of light increases kinetic energy of photoelectrons.
  Avoid: Kinetic energy depends only on frequency; intensity affects only the number of photoelectrons.
• Trap: Thinking photoelectric emission occurs for any frequency if light is intense enough.
  Avoid: Emission occurs only if frequency ≥ threshold frequency, regardless of intensity.
• Trap: Confusing stopping potential with accelerating potential.
  Avoid: Stopping potential opposes electron motion and measures ( K_{\text{max}} ); it is always negative in sign convention but magnitude used.

Practice MCQs

1. Question: Which of the following best describes the dependence of maximum kinetic energy of photoelectrons on the frequency of incident light?
   
   A. It decreases linearly with frequency
   
   B. It increases parabolically with frequency
   
   C. It increases linearly with frequency above threshold
   
   D. It is independent of frequency
   Answer: C
   Explanation: ( K_{\text{max}} = h\nu - \phi_0 ), so it increases linearly with ( \nu ) above ( \nu_0 ).
   Why others fail: Option B is tempting if confusing with classical energy predictions.

2. Question: The work function of a metal is 3.2 eV. What is the threshold frequency?
   
   A. ( 4.8 \times 10^{14} \, \text{Hz} )
   
   B. ( 7.7 \times 10^{14} \, \text{Hz} )
   
   C. ( 3.2 \times 10^{15} \, \text{Hz} )
   
   D. ( 1.6 \times 10^{14} \, \text{Hz} )
   Answer: B
   Explanation: ( \nu_0 = \frac{\phi_0}{h} = \frac{3.2 \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 7.7 \times 10^{14} \, \text{Hz} ).
   Why others fail: Option A is common error from using ( h \approx 6.6 \times 10^{-34} ) without proper conversion of eV to joules.

3. Question: A graph of stopping potential ( V_0 ) versus frequency ( \nu ) of incident light is plotted for two metals X and Y. If the slope for X is greater than Y, which statement is correct?
   
   A. Work function of X is higher
   
   B. Threshold frequency of Y is lower
   
   C. Slope is same for all metals
   
   D. X has higher photoelectric sensitivity
   Answer: C
   Explanation: Slope of ( V_0 ) vs ( \nu ) is ( h/e ), a universal constant, so it must be same for all metals.
   Why others fail: Students often assume different slopes due to different work functions, but slope is independent of material.

4. Question: If the frequency of incident radiation is doubled, what happens to the stopping potential?
   
   A. Doubles
   
   B. Becomes more than double
   
   C. Increases by a fixed amount depending on work function
   
   D. Remains unchanged
   Answer: C
   Explanation: ( V_0 = \frac{h\nu - \phi_0}{e} ), so change depends on initial ( \nu ) and ( \phi_0 ); not a fixed ratio.
   Why others fail: Option A is tempting due to linear equation, but intercept (( \phi_0 )) prevents doubling.

5. Question: Light of wavelength 300 nm is incident on a metal with work function 2.5 eV. What is the maximum kinetic energy of emitted electrons?
   
   A. 1.6 eV
   
   B. 2.6 eV
   
   C. 4.1 eV
   
   D. 6.6 eV
   Answer: A
   Explanation: ( E = \frac{1240}{300} \approx 4.13 \, \text{eV} ), so ( K_{\text{max}} = 4.13 - 2.5 = 1.63 \approx 1.6 \, \text{eV} ).
   Why others fail: Option C is the photon energy, tempting if forgetting to subtract work function.

Last‑Minute Revision

• ⚠️ Einstein’s equation: ( K_{\text{max}} = h\nu - \phi_0 ) — must remember form.
• ⚠️ Threshold frequency ( \nu_0 = \phi_0 / h ) — minimum frequency for emission.
• ⚠️ No emission if ( \nu < \nu_0 ), even with high intensity.
• ⚠️ ( K_{\text{max}} ) depends only on ( \nu ), not intensity.
• ⚠️ Stopping potential ( V_0 = (h\nu - \phi_0)/e ).
• ⚠️ Slope of ( V_0 ) vs ( \nu ) graph = ( h/e ) — same for all metals.
• ⚠️ Intercept on frequency axis = ( \nu_0 ); on voltage axis = ( -\phi_0/e ).
• ⚠️ One photon ejects one electron — 1:1 interaction.
• ⚠️ Work function in eV → convert to joules using ( 1.6 \times 10^{-19} ).
• ⚠️ Photon energy ( E = hc/\lambda ); use ( hc = 1240 \, \text{eV·nm} ) for quick calculations.
• ⚠️ Cesium has low work function (~2.14 eV) — used in photocells.
• ⚠️ Photoelectric effect proves particle nature of light.
• ⚠️ Time delay in emission is negligible (< 10⁻⁹ s).
• ⚠️ Intensity ∝ number of photons ∝ photocurrent.
• ⚠️ Nobel Prize to Einstein (1921) — for photoelectric effect, not relativity.
• ⚠️ Cut-off wavelength ( \lambda_0 = c / \nu_0 = hc / \phi_0 ).
• ⚠️ If ( \nu = \nu_0 ), then ( K_{\text{max}} = 0 ), ( V_0 = 0 ).
• ⚠️ Use ( h = 6.626 \times 10^{-34} \, \text{J s} ), ( e = 1.6 \times 10^{-19} \, \text{C} ).
• ⚠️ Mnemonic: “Frequency Freezes Kinetic energy” — only frequency controls ( K_{\text{max}} ).