Q.Consider a beam of electrons (each electron with energy $E_0$) incident on a metal surface kept in an evacuated chamber. Then

Showing the 12 most recent of 40 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.When a metal surface is illuminated with lights of wavelengths $\lambda$ and $2\lambda$ separately, the stopping potentials are V and $\dfrac{V}{3}$ respectively. Then the threshold wavelength of that metal surface is (A) $\dfrac{4\lambda}{3}$ (B) $4\lambda$ (C) $6\lambda$ (D) $\dfrac{8\lambda}{3}$

   ›Reveal solutionSolution

   Using Einstein's photoelectric equation at two different wavelengths (and their given stopping potentials) as simultaneous equations lets us solve for the work function, and hence the threshold wavelength. Answer: $4\lambda$.

   Concept and Intuition

   Einstein's photoelectric equation, $eV_s = \dfrac{hc}{\lambda} - \phi$, relates the stopping potential to the photon energy and the material's work function $\phi$. Since $\phi$ is a fixed property of the metal, two different (wavelength, stopping potential) pairs give two equations that can be solved simultaneously for $\phi$, and then the threshold wavelength follows from $\phi = hc/\lambda_0$.

   Step-by-Step Solution

   1. For wavelength $\lambda$: $eV = \dfrac{hc}{\lambda} - \phi$ … (i)
   2. For wavelength $2\lambda$: $e\dfrac{V}{3} = \dfrac{hc}{2\lambda} - \phi$ … (ii)
   3. From (i): $\phi = \dfrac{hc}{\lambda} - eV$. Substitute into (ii): $\dfrac{eV}{3} = \dfrac{hc}{2\lambda} - \dfrac{hc}{\lambda} + eV$.
   4. Simplify: $\dfrac{eV}{3} - eV = \dfrac{hc}{2\lambda} - \dfrac{hc}{\lambda} \Rightarrow -\dfrac{2eV}{3} = -\dfrac{hc}{2\lambda} \Rightarrow eV = \dfrac{3hc}{4\lambda}$.
   5. Then $\phi = \dfrac{hc}{\lambda} - \dfrac{3hc}{4\lambda} = \dfrac{hc}{4\lambda}$.
   6. Threshold wavelength: $\lambda_0 = \dfrac{hc}{\phi} = \dfrac{hc}{hc/(4\lambda)} = 4\lambda$.

   Common Mistakes

   • Sign errors when substituting $\phi$ back into the second equation.
   • Confusing threshold wavelength ($\phi = hc/\lambda_0$) with threshold frequency.

   ✓Final answer

   The correct option is (B) — $4\lambda$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Two identical photo cathodes receive light of frequencies $f_1$ and $f_2$. If the velocity of photo electrons (of mass m) coming out are respectively $V_1$ & $V_2$, then (A) $V_1^2-V_2^2=\dfrac{2h}{m}(f_1-f_2)$ (B) $V_1+V_2=\left[\dfrac{2h}{m}(f_1+f_2)\right]^{1/2}$ (C) $V_1^2+V_2^2=\dfrac{2h}{m}(f_1+f_2)$ (D) $V_1-V_2=\left[\dfrac{2h}{m}(f_1-f_2)\right]^{1/2}$

   ›Reveal solutionSolution

   Writing Einstein's photoelectric equation for each frequency and subtracting to cancel the common work function directly gives the relation among $V_1,V_2,f_1,f_2$. Answer: (A).

   Concept and Intuition

   Einstein's photoelectric equation states the maximum kinetic energy of an emitted photoelectron is $KE_{max}=hf-\phi_0$, where $\phi_0$ is the cathode's work function. Since both cathodes are identical, $\phi_0$ is the same for both — so subtracting the two equations eliminates the unknown work function entirely, leaving a clean relation between the observed velocities and frequencies.

   Step-by-Step Solution

   1. For frequency $f_1$: $\tfrac12 m V_1^2 = hf_1-\phi_0$.
   2. For frequency $f_2$: $\tfrac12 m V_2^2 = hf_2-\phi_0$.
   3. Subtract (2) from (1): $\tfrac12 m(V_1^2-V_2^2)=h(f_1-f_2)$.
   4. Multiply both sides by $2/m$: $V_1^2-V_2^2=\dfrac{2h}{m}(f_1-f_2)$.

   Common Mistakes

   • Trying to eliminate $\phi_0$ by adding the equations instead of subtracting — that leaves $\phi_0$ in the result.
   • Taking the square root prematurely and writing a relation in $V_1-V_2$ (that's not how the algebra works here; $V_1^2-V_2^2$ is the quantity that cleanly cancels $\phi_0$).

   ✓Final answer

   The correct option is (A) — $V_1^2-V_2^2=\dfrac{2h}{m}(f_1-f_2)$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The kinetic energy of released electron in photo electric effect depends on (A) Intensity of incidenting photons (B) Frequency of incidenting photons (C) Area of photocell (D) Time

   ›Reveal solutionSolution

   Einstein's photoelectric equation shows KE of ejected electrons depends on photon frequency, not intensity. Answer: frequency of incident photons.

   Concept and Intuition

   The photoelectric effect was the key evidence for the particle (photon) nature of light. Each photon interacts with one electron, transferring its full energy $h\nu$; some of that energy overcomes the work function $\phi$ of the metal, and the rest becomes the electron's kinetic energy: $KE_{max} = h\nu - \phi$. Since this equation has no dependence on how many photons arrive per second (intensity), increasing intensity only increases the number of ejected electrons (photocurrent), not their individual kinetic energy. Only increasing the frequency increases each photon's energy and hence the ejected electron's energy.

   Step-by-Step Solution

   1. Einstein's photoelectric equation: $KE_{max} = h\nu - \phi$, where $\phi$ is the material's work function (fixed for a given metal).
   2. $KE_{max}$ depends explicitly and only on $\nu$ (and the fixed $\phi$) — no intensity, area, or time term appears.
   3. Intensity affects the number of photons per second, which affects photocurrent (number of electrons/sec), not the energy per electron.
   4. Hence the correct dependence is on frequency.

   Common Mistakes

   • Assuming brighter (more intense) light means 'more energetic' electrons — intensity only means more photons, not higher-energy photons.
   • Confusing this with photocurrent magnitude, which does depend on intensity.

   ✓Final answer

   The correct option is (B) — Frequency of incidenting photons.

   ANSWER: B

4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Sodium and Copper have work functions 2.3 eV and 4.5 eV respectively. Then the ratio of their threshold wavelengths is nearly (A) $1:2$ (B) $4:1$ (C) $2:1$ (D) $1:4$

   ›Reveal solutionSolution

   Threshold wavelength $\propto 1/\phi$, so the ratio of Na's to Cu's threshold wavelength is $\phi_{Cu}/\phi_{Na}=4.5/2.3\approx 2:1$.

   Concept and Intuition

   The photoelectric threshold condition is $h\nu_{th}=\phi$, i.e. $\dfrac{hc}{\lambda_{th}}=\phi$, giving $\lambda_{th}=\dfrac{hc}{\phi}$. A metal with a smaller work function has a longer threshold wavelength (easier to eject electrons, so even lower-energy/longer-wavelength photons suffice). Sodium has the smaller work function of the two, so it should have the longer threshold wavelength.

   Step-by-Step Solution

   1. $\lambda_{th,Na} = \dfrac{hc}{\phi_{Na}}$, $\lambda_{th,Cu} = \dfrac{hc}{\phi_{Cu}}$.
   2. Ratio: $\dfrac{\lambda_{th,Na}}{\lambda_{th,Cu}} = \dfrac{\phi_{Cu}}{\phi_{Na}} = \dfrac{4.5}{2.3} = 1.956$.
   3. This rounds to nearly $2:1$ — sodium's threshold wavelength is about twice copper's, consistent with sodium being much easier to photo-ionize (smaller work function).

   Common Mistakes

   • Taking the ratio the wrong way round ($\phi_{Na}/\phi_{Cu}$ instead of the reciprocal), which would wrongly suggest Cu has the longer threshold wavelength.
   • Forgetting that $\lambda_{th}$ and $\phi$ are inversely, not directly, proportional.

   ✓Final answer

   The correct option is (C) — $2:1$.

   ANSWER: C

5. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.In photo electric experiment, if the wavelength of incident light on the metal changes from 200 nm to 300 nm. The decrease in stopping potential is about $\left(\dfrac{hc}{e} = 1240\ eV\text{-}nm\right)$ (A) 2.1 V (B) 4.2 V (C) 3.1 V (D) 6.2 V

   ›Reveal solutionSolution

   Since work function cancels when comparing two stopping potentials for the same metal, the drop in stopping potential is just $\dfrac{hc}{e}\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_2}\right)\approx2.1$ V.

   Concept and Intuition

   Einstein's photoelectric equation: $eV_s=\dfrac{hc}{\lambda}-\phi_0$, where $\phi_0$ is the (fixed) work function of the metal. Since the same metal is used for both wavelengths, $\phi_0$ is common and cancels out when we take the difference of stopping potentials at two wavelengths — leaving only the photon energy difference.

   Step-by-Step Solution

   1. At $\lambda_1=200$ nm: $eV_{s1}=\dfrac{hc}{\lambda_1}-\phi_0$.
   2. At $\lambda_2=300$ nm: $eV_{s2}=\dfrac{hc}{\lambda_2}-\phi_0$.
   3. Decrease in stopping potential: $e(V_{s1}-V_{s2})=hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_2}\right)$.
   4. Using $\dfrac{hc}{e}=1240$ eV·nm: $\Delta V_s=1240\left(\dfrac{1}{200}-\dfrac{1}{300}\right)=1240\times\dfrac{3-2}{600}=1240/600\approx2.07\ \text{V}$.
   5. Rounding, $\Delta V_s\approx2.1$ V.

   Common Mistakes

   • Trying to compute the work function separately (not given, and not needed since it cancels).
   • Arithmetic slip in combining $1/200-1/300$ as a single fraction.

   ✓Final answer

   The correct option is (A) — 2.1 V.

   ANSWER: A

6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the threshold wavelength of light for photoelectric emission to take place from a metal surface is 6250 Å, then the work function of the metal is (Planck's constant = $6.6\times10^{-34}$ Js) (A) 3.98 eV (B) 1.98 eV (C) 2.98 eV (D) 4.98 eV

   ›Reveal solutionSolution

   This tests the relation between threshold wavelength and work function in the photoelectric effect. The work function comes out to 1.98 eV.

   Concept and Intuition

   The threshold wavelength is the longest wavelength (lowest photon energy) that can just barely eject an electron from the metal, with zero kinetic energy left over. At exactly this wavelength, the entire photon energy $hc/\lambda_0$ equals the work function $\phi$ — the minimum energy needed to free an electron from the metal surface.

   Step-by-Step Solution

   1. At threshold: $\phi=\dfrac{hc}{\lambda_0}$.
   2. Substitute values: $h=6.6\times10^{-34}$ Js, $c=3\times10^8$ m/s, $\lambda_0=6250\ \text{Å}=6250\times10^{-10}\ \text{m}=6.25\times10^{-7}$ m.

   $\phi=\frac{(6.6\times10^{-34})(3\times10^8)}{6.25\times10^{-7}}=\frac{1.98\times10^{-25}}{6.25\times10^{-7}}=3.168\times10^{-19}\ \text{J}$

   3. Convert to eV (divide by $1.6\times10^{-19}$ J/eV):

   $\phi=\frac{3.168\times10^{-19}}{1.6\times10^{-19}}\approx1.98\ \text{eV}$

   Common Mistakes

   • Forgetting to convert the wavelength from Å to metres before using it (a factor-of-$10^{10}$ error).
   • Forgetting the final J-to-eV conversion, leaving the answer in joules.

   ✓Final answer

   The correct option is (B) — 1.98 eV.

   ANSWER: B

7. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.In a photoelectric experiment, when the wavelength of light incident on a metal is $\lambda$, the maximum kinetic energy of the emitted photoelectron is E. When the wavelength of incident light is $\dfrac{\lambda}{3}$, the maximum kinetic energy of the emitted photoelectron becomes 4E. The work function of the metal is (A) $\dfrac{hc}{\lambda}$ (B) $\dfrac{3hc}{\lambda}$ (C) $\dfrac{hc}{2\lambda}$ (D) $\dfrac{hc}{3\lambda}$

   ›Reveal solutionSolution

   Setting up Einstein's photoelectric equation for two different wavelengths and solving the resulting pair of linear equations gives the work function as $\dfrac{hc}{3\lambda}$.

   Concept and Intuition

   Einstein's photoelectric equation, $KE_{max}=\dfrac{hc}{\lambda}-\phi$, is linear in $\dfrac1\lambda$ and $\phi$. Given two (wavelength, KE) pairs, we get two equations in two unknowns ($E$ appears as a scale, $\phi$ is what we want) which we can solve algebraically.

   Step-by-Step Solution

   1. First case: $\dfrac{hc}{\lambda}-\phi = E$ → so $\dfrac{hc}{\lambda}=E+\phi$.
   2. Second case (wavelength $\lambda/3$): $\dfrac{hc}{\lambda/3}-\phi=4E \Rightarrow \dfrac{3hc}{\lambda}-\phi=4E$.
   3. Substitute $\dfrac{hc}{\lambda}=E+\phi$ into the second equation: $3(E+\phi)-\phi=4E \Rightarrow 3E+2\phi=4E \Rightarrow \phi=\dfrac{E}{2}$.
   4. From step 1: $\dfrac{hc}{\lambda}=E+\dfrac{E}{2}=\dfrac{3E}{2} \Rightarrow E=\dfrac{2}{3}\cdot\dfrac{hc}{\lambda}$.
   5. So $\phi=\dfrac{E}{2}=\dfrac{1}{3}\cdot\dfrac{hc}{\lambda}=\dfrac{hc}{3\lambda}$.

   Common Mistakes

   • Treating $E$ as a known numeric constant rather than eliminating it algebraically.
   • Sign errors when substituting the two equations.

   ✓Final answer

   The correct option is (D) — $\dfrac{hc}{3\lambda}$.

   ANSWER: D

8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.When photons of energy $8\times10^{-19}\ \mathrm{J}$ incident on a photosensitive material, the de Broglie wavelength of the photoelectrons emitted with maximum kinetic energy is 10 Å. The work function of the photosensitive material is nearly (A) 3.5 eV (B) 2.5 eV (C) 2.0 eV (D) 1.5 eV

   ›Reveal solutionSolution

   This tests Einstein's photoelectric equation combined with the de Broglie relation; the photon energy is 5 eV, the photoelectron's kinetic energy (from its de Broglie wavelength) is about 1.5 eV, so the work function is 3.5 eV.

   Concept and Intuition

   Einstein's photoelectric equation, $E_{photon} = \phi + KE_{max}$, says the absorbed photon's energy splits between freeing the electron (the work function $\phi$) and giving it kinetic energy. Here the electron's maximum kinetic energy isn't given directly — instead its de Broglie wavelength is, so we must first recover $KE_{max}$ from $\lambda = h/p$ and $KE=p^2/2m$, then use the photoelectric equation to isolate $\phi$.

   Step-by-Step Solution

   1. Photon energy in eV: $E = \dfrac{8\times10^{-19}}{1.6\times10^{-19}} = 5\,\mathrm{eV}$.
   2. Momentum of the photoelectron from its de Broglie wavelength: $p = \dfrac{h}{\lambda} = \dfrac{6.63\times10^{-34}}{10\times10^{-10}} = 6.63\times10^{-25}\,\mathrm{kg\,m/s}$.
   3. Kinetic energy: $KE_{max} = \dfrac{p^2}{2m_e} = \dfrac{(6.63\times10^{-25})^2}{2\times9.11\times10^{-31}} \approx 2.41\times10^{-19}\,\mathrm{J} \approx 1.5\,\mathrm{eV}$.
   4. Work function: $\phi = E_{photon} - KE_{max} = 5 - 1.5 = 3.5\,\mathrm{eV}$.

   Common Mistakes

   • Trying to use $\lambda = h/p$ with the photon's energy instead of computing the electron's momentum from its own de Broglie wavelength — the two are different particles here.
   • Forgetting to convert the final kinetic energy from joules to eV before subtracting from the photon energy (also in eV).

   ✓Final answer

   The correct option is (A) — 3.5 eV.

   ANSWER: A

9. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.When a metal surface is illuminated by a light of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated by light of wavelength $2\lambda$, the stopping potential is $\dfrac{V}{4}$, the threshold wavelength is (A) $\lambda$ (B) $2\lambda$ (C) $3\lambda$ (D) $\dfrac{\lambda}{2}$

   ›Reveal solutionSolution

   Applying Einstein's photoelectric equation at two wavelengths and eliminating the work function gives threshold wavelength $\lambda_0 = 3\lambda$.

   Concept and Intuition

   The stopping potential $V$ measures the maximum kinetic energy of photoelectrons: $eV = \dfrac{hc}{\lambda} - \phi$, where $\phi = \dfrac{hc}{\lambda_0}$ is the work function expressed via the threshold wavelength $\lambda_0$. Given the stopping potential at two different wavelengths, we get two linear equations in the two unknowns $eV$ and $\dfrac{hc}{\lambda_0}$ (treating $\dfrac{hc}{\lambda}$ as known), which can be solved simultaneously.

   Step-by-Step Solution

   1. At wavelength $\lambda$: $eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$ ... (1)
   2. At wavelength $2\lambda$: $e\cdot\dfrac{V}{4} = \dfrac{hc}{2\lambda} - \dfrac{hc}{\lambda_0}$ ... (2)
   3. Subtract (2) from (1): $eV - \dfrac{eV}{4} = \dfrac{hc}{\lambda} - \dfrac{hc}{2\lambda}$, i.e. $\dfrac{3eV}{4} = \dfrac{hc}{2\lambda}$, so $eV = \dfrac{2}{3}\cdot\dfrac{hc}{\lambda}$.
   4. Substitute into (1): $\dfrac{2}{3}\cdot\dfrac{hc}{\lambda} = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$, so $\dfrac{hc}{\lambda_0} = \dfrac{1}{3}\cdot\dfrac{hc}{\lambda}$.
   5. Therefore $\lambda_0 = 3\lambda$.

   Common Mistakes

   • Forgetting that stopping potential scales with $V$, not $eV$ directly when substituting the fraction $V/4$ — must multiply through by $e$ consistently.
   • Sign errors when subtracting the two photoelectric equations.

   ✓Final answer

   The correct option is (C) — $3\lambda$.

   ANSWER: C

10. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.The work function of a photosensitive metal surface is $6.4 \times 10^{-19}$ J. The maximum kinetic energy of the emitted photoelectrons when electromagnetic radiation of wavelength 1240 Å incidents on the metal surface is nearly (A) 5 eV (B) 6 eV (C) 3 eV (D) 4 eV

    ›Reveal solutionSolution

    Converting the work function to eV and the photon energy via $E=1240/\lambda(\text{nm})$ gives a maximum photoelectron kinetic energy of 6 eV.

    Concept and Intuition

    Einstein's photoelectric equation states $KE_{max}=E_{photon}-\phi$, where $\phi$ is the work function (minimum energy needed to free an electron) and $E_{photon}=\dfrac{hc}{\lambda}$ is the incident photon's energy. A handy shortcut is $E_{photon}(\text{eV})=\dfrac{1240}{\lambda(\text{nm})}$, since $hc\approx 1240$ eV·nm.

    Step-by-Step Solution

    1. Convert work function to eV: $\phi=\dfrac{6.4\times10^{-19}}{1.6\times10^{-19}}=4$ eV.
    2. Convert wavelength to nm: $1240\ \text{Å}=124$ nm.
    3. Photon energy: $E=\dfrac{1240}{124}=10$ eV.
    4. Maximum kinetic energy: $KE_{max}=E-\phi=10-4=6$ eV.

    Common Mistakes

    • Mixing up units (Å vs nm) when applying the $1240/\lambda(\text{nm})$ shortcut — 1240 Å must first become 124 nm.
    • Forgetting to convert the work function from joules to eV before subtracting.

    ✓Final answer

    The correct option is (B) — 6 eV.

    ANSWER: B

11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The longest wavelength of light that can initiate photo electric effect in the metal of work function 9 eV is (A) $1.37 \times 10^{-7}$ m (B) $1.5 \times 10^{-7}$ m (C) $3.7 \times 10^{-7}$ m (D) $4 \times 10^{-7}$ m

    ›Reveal solutionSolution

    This tests the threshold-wavelength relation in the photoelectric effect: the longest wavelength that can still eject an electron corresponds exactly to a photon energy equal to the work function.

    Concept and Intuition

    Below the threshold frequency (equivalently, above the threshold/longest wavelength), photons don't carry enough energy to overcome the metal's work function, so no photoelectrons are emitted regardless of intensity. At exactly the threshold, the photon energy just equals the work function, with zero kinetic energy left over for the electron.

    Step-by-Step Solution

    1. At the threshold, $E_{photon}=\dfrac{hc}{\lambda_{max}}=W$ (work function), since KE of the emitted electron is zero here.
    2. So $\lambda_{max}=\dfrac{hc}{W}$.
    3. Using $hc=1240$ eV·nm and $W=9$ eV: $\lambda_{max}=\dfrac{1240\ \text{eV·nm}}{9\ \text{eV}}\approx137.8$ nm.
    4. Converting to metres: $137.8\text{ nm}=1.378\times10^{-7}\text{ m}\approx1.37\times10^{-7}$ m.

    Common Mistakes

    • Using $hc=1240$ eV·nm but forgetting to convert nm to metres at the end.
    • Confusing threshold wavelength (maximum wavelength that still works) with a minimum wavelength condition.

    ✓Final answer

    The correct option is (A) — $1.37\times10^{-7}$ m.

    ANSWER: A

12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.Light of wavelength $4000\,\text{\AA}$ is incident on a sodium surface for which the threshold wavelength of photo electrons is $5420\,\text{\AA}$. The work function of sodium is (A) 4.58 eV (B) 2.29 eV (C) 1.14 eV (D) 0.57 eV

    ›Reveal solutionSolution

    The work function equals the photon energy corresponding to the threshold wavelength, computed using $hc\approx12400\ \text{eV·Å}$.

    Concept and Intuition

    In the photoelectric effect, the threshold wavelength $\lambda_0$ is the longest wavelength (lowest photon energy) that can still just eject an electron from the metal surface — at this wavelength the photon energy exactly equals the work function, with zero kinetic energy left over for the electron. The incident wavelength of 4000 Å given in the problem is a distractor for this particular sub-question, since the work function depends only on the threshold wavelength.

    Step-by-Step Solution

    1. Work function: $W = \dfrac{hc}{\lambda_0}$.
    2. Using the convenient constant $hc \approx 12400\ \text{eV·Å}$ and $\lambda_0 = 5420$ Å:
    3. $W = \dfrac{12400}{5420} \approx 2.288\ \text{eV} \approx 2.29$ eV.

    Common Mistakes

    • Using the incident wavelength (4000 Å) instead of the threshold wavelength (5420 Å) to compute the work function.
    • Using an imprecise value of $hc$ leading to a different rounding than the intended answer.

    ✓Final answer

    The correct option is (B) — 2.29 eV.

    ANSWER: B