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In a photoelectric effect experiment, light of frequencies $f_1 > f_2 > f_3$ is shone on a metal. It is observed that $f_3$ is the threshold frequency. Which of the following correctly describes the stopping potentials $V_1$, $V_2$, $V_3$?

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Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
$V_1 = V_2 = V_3$
2
$V_1 > V_2 > V_3 = 0$
3
$V_1 < V_2 < V_3$
4
$V_1 > V_2$, $V_3 = 0$
Correct Answer
$V_1 > V_2$, $V_3 = 0$
Solution
1

$V = h(f - f_0)/e$; $f_3 = f_0$ (threshold)

$V_3 = h(f_0 - f_0)/e = 0$

2

$V_1 = h(f_1 - f_0)/e > 0$, $V_2 = h(f_2 - f_0)/e > 0$, $V_1 > V_2$ (since $f_1 > f_2$)

Answer: $V_1 > V_2$, $V_3 = 0$

Stopping potential $V = h(f-f_0)/e$; threshold $f_3=f_0 \Rightarrow V_3=0$
$V_1>V_2>V_3=0$; stopping potential depends on frequency, NOT intensity
Theory: Modern Physics / Photoelectric Effect
1. Photoelectric Effect

Photoelectric effect: emission of electrons from metal surface when light of sufficient frequency is incident. Discovered by Hertz (1887), explained by Einstein (1905). Key observations: threshold frequency exists ($f_0$), instantaneous emission, $KE_{max}$ depends on $f$ not intensity, photocurrent proportional to intensity. Einstein\'s explanation: light comes in quanta (photons) each with energy $E = hf$. One photon ejects one electron if $hf > \phi$.

2. Photon Properties

Photon: quantum of electromagnetic radiation. Energy: $E = hf = hc/\lambda$. Momentum: $p = E/c = h/\lambda = hf/c$. Mass: $m = 0$ (rest mass), but has relativistic mass $m = E/c^2$. Photon is its own antiparticle. Speed always $c$ in vacuum. Planck\'s constant: $h = 6.626\times10^{-34}$ J s. Compton effect: photon-electron collision, photon wavelength increases: $\Delta\lambda = (h/m_ec)(1-\cos\phi)$ (Compton wavelength shift).

3. de Broglie Hypothesis

de Broglie (1924): particles have wave nature. de Broglie wavelength: $\lambda = h/p = h/mv$. For electron accelerated through potential $V$: $\lambda = h/\sqrt{2meV}$. Davisson-Germer experiment (1927): confirmed electron diffraction, proving wave nature. Heisenberg uncertainty: $\Delta x \cdot \Delta p \geq h/4\pi$, $\Delta E \cdot \Delta t \geq h/4\pi$.

4. Bohr Model of Hydrogen

Bohr (1913): electrons move in circular orbits with quantised angular momentum $mvr = nh/2\pi$. Energy levels: $E_n = -13.6/n^2$ eV. Radius: $r_n = 0.529 n^2$ Angstrom. Transition: $hf = E_i - E_f$. Spectral series: Lyman (UV, $n_f=1$), Balmer (visible, $n_f=2$), Paschen (IR, $n_f=3$), Brackett ($n_f=4$), Pfund ($n_f=5$). Limitations: cannot explain multi-electron atoms, intensity of spectral lines, fine structure.

Frequently Asked Questions
1. What is stopping potential in photoelectric effect? ⌄
Stopping potential $V_0$ is the minimum negative voltage applied to the collector electrode that just stops the photoelectric current. It measures the maximum KE of ejected electrons: $eV_0 = KE_{max} = h(f - f_0)$ where $f_0$ is threshold frequency, $h$ is Planck\'s constant.
2. What is threshold frequency? ⌄
Threshold frequency $f_0$ is the minimum frequency of light that can cause photoelectric emission. Below $f_0$: no emission regardless of intensity. At $f_0$: electrons emitted with zero KE (stopping potential = 0). Above $f_0$: electrons emitted with $KE = h(f - f_0)$.
3. Does stopping potential depend on intensity? ⌄
No! Stopping potential depends ONLY on frequency (energy of photons), NOT on intensity. Intensity affects: number of photoelectrons emitted (photocurrent). More intensity = more photons = more electrons emitted (higher current) but NOT higher energy per electron. This was the key experimental observation that disproved classical wave theory.
4. What was Einstein's photoelectric equation? ⌄
Einstein (1905, Nobel Prize 1921): $KE_{max} = hf - \phi$ where $\phi = hf_0$ is the work function (minimum energy needed to eject electron). $eV_0 = hf - hf_0$. Graph of $V_0$ vs $f$ is linear with slope $h/e$ (allows measurement of Planck\'s constant).
5. What are the experimental observations of photoelectric effect? ⌄
1. No emission below threshold frequency. 2. Emission is instantaneous (no time delay). 3. Maximum KE increases with frequency (not intensity). 4. Photocurrent increases with intensity. These observations cannot be explained by classical wave theory (which predicted: any frequency if intensity high enough, time delay for low intensity, KE increasing with intensity). Einstein's photon concept explained all these correctly.
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