Q.How much positive and negative charge is there in a cup of water? (Assume the mass of one cup of water to be $250\,\text{g}$.)

Showing the 12 most recent of 22 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The variation of the stopping potential ( $\mathrm{V}_{\mathrm{o}}$ ) with the frequency of incident radiation( $n$ ) is as given below. [no - threshold frequency. h - Planck's constant e-electronic charge] then the slope of the graph AB with the frequency axis is: (A) $\frac{h}{e}$ (B) $\frac{h n_0}{V_0}$ (C) $\frac{n_0}{V_0}$ (D) $\frac{V_0}{n_0}$

   ›Reveal solutionSolution

   The slope of the stopping potential vs. frequency graph is given by the photoelectric equation $eV_o = h(n - n_0)$, so the slope is $\frac{h}{e}$. The correct option is (A).

   The key concept here is the photoelectric effect equation as formulated by Einstein. The stopping potential $V_o$ is the voltage that just stops the most energetic photoelectrons. The energy of those electrons equals the photon energy minus the work function. When you plot $V_o$ against frequency $n$, the graph is a straight line whose slope is a fundamental constant ratio.

   Why this approach works:

   The equation $eV_o = h n - \phi$ (where $\phi = h n_0$ is the work function) is linear in $n$. The slope of the $V_o$ vs. $n$ line is therefore $h/e$. The graph’s intercept on the frequency axis is the threshold frequency $n_0$. The dashed triangle in the figure simply illustrates that slope = $\Delta V_o / \Delta n$, which must equal $h/e$.

   Let’s work through it step by step:

   1. Write the photoelectric equation. Einstein’s photoelectric equation for the stopping potential is:

   $eV_o = h n - \phi$

   where $e$ is the electron charge, $h$ is Planck’s constant, $n$ is the frequency of incident radiation, and $\phi$ is the work function of the metal.

   2. Express the work function in terms of threshold frequency. The threshold frequency $n_0$ is the minimum frequency needed to eject electrons, so at $n = n_0$, $V_o = 0$. Substituting:

   $0 = h n_0 - \phi \quad \Rightarrow \quad \phi = h n_0$

   Thus the equation becomes:

   $eV_o = h n - h n_0 = h (n - n_0)$

   3. Solve for $V_o$ to see the linear relationship. Divide both sides by $e$:

   $V_o = \frac{h}{e} (n - n_0)$

   This is of the form $V_o = m (n - n_0)$, where $m$ is the slope.

   4. Identify the slope from the graph. The graph shows $V_o$ on the vertical axis and $n$ on the horizontal axis. The line AB has a constant slope. From the equation, the slope is the coefficient of $n$: $\text{slope} = \frac{h}{e}$ …

   🔒 Start your 14-day free trial to see the full solution →
2. KCET 2026Set C21 markMCQ
   Q.Work function of the metal is (A) Maximum possible energy acquired by an electron (B) Equal for all metals (C) Minimum energy required by an electron to just eject from metal surface (D) Maximum energy which is given to electron to move out of metal surface

   ›Reveal solutionSolution

   The work function $\phi_0$ is a property of the metal defined as the minimum energy needed to remove (eject) a free electron from just inside its surface.

   Step 1 — Physical picture

   Electrons in a metal are bound by attractive forces from the positive ion lattice. To escape the metal surface entirely, an electron must be given at least enough energy to overcome this binding — no less will do, since a smaller energy leaves the electron still trapped.

   Step 2 — Definition of work function …

   🔒 Start your 14-day free trial to see the full solution →
3. KCET 2026Set C21 markMCQ
   Q.Variation of photoelectric current with anode potential is shown below . Choose the correct option ($V_0$ = stopping potential). (A) Graph-I: the current is zero at the negative anode potential -V0, rises smoothly through the origin as the anode potential increases, and then levels off at a constant (saturation) value at higher positive potentials (B) Graph-II: the current is zero at -V0 and rises in a straight line with anode potential, continuing to increase without ever levelling off (C) Graph-III: the current is at its highest value at -V0 and falls in a straight line down to zero near the origin, i.e. current decreases as anode potential increases (D) Graph-IV: the current starts at zero at the origin itself (no negative stopping-potential intercept is shown) and increases with an upward-curving (concave-up) shape as the anode potential increases

   ›Reveal solutionSolution

   Photoelectric current falls to exactly zero at the stopping potential $-V_0$ (a negative potential, not the origin), rises smoothly (not abruptly or linearly forever) as the anode potential increases, and levels off at a constant saturation current once every emitted photoelectron is being collected.

   Step 1 — Behaviour at negative (retarding) anode potential

   As the anode potential is made increasingly negative, it opposes the motion of the photoelectrons emitted with kinetic energy up to $eV_0$. The current decreases continuously and becomes exactly zero only when the anode potential reaches $-V_0$, the stopping potential — not at the origin, and not abruptly.

   Step 2 — Behaviour as the anode potential increases through zero and beyond

   As the anode potential increases from $-V_0$ through zero and into positive values, more and more of the emitted photoelectrons (which have a spread of kinetic energies from $0$ up to $eV_0$) are attracted to and collected by the anode, so the current rises smoothly (not as a straight line, and not with an upward-curving shape).

   Step 3 — Saturation …

   🔒 Start your 14-day free trial to see the full solution →
4. KCET 2025Set D-41 markMCQ
   Q.If AB is incident plane wave front then refracted wave front in $(n_2 > n_1)$ (A) [FIGURE] (B) [FIGURE] (C) [FIGURE] (D) [FIGURE]

   ›Reveal solutionSolution

   The lens is thicker at the centre and slows light more there, so the wavefront's middle lags: the plane wavefront emerges as a converging spherical wavefront centred on the focus.

   Step 1 — Huygens' picture of refraction through a lens.

   Inside the lens the wave travels slower: $v_2=c/n_2 < v_1=c/n_1$ since $n_2>n_1$. The convex lens is thickest at its centre, so the central part of the incident plane wavefront spends the longest time in the slow medium and is retarded most, while the edges (thin glass) get through with least delay.

   Step 2 — What that does to the shape.

   The centre of the wavefront falls behind the edges, so the flat front $AB$ emerges curved, with its middle bent backwards — that is, the emerging wavefront is a spherical surface whose centre of curvature lies ahead of it (to the right), at the focus $F$. Drawn on paper, such a converging wavefront looks like a "(" — bulging towards the left, concave towards the right.

   Step 3 — Cross-check with the rays. …

   🔒 Start your 14-day free trial to see the full solution →
5. COMEDK 2025Set 2025-A1 markMCQ
   Q.Which of the following statement is wrong regarding the photo electric effect. (A) Photo electric current is directly proportional to the intensity of the incident radiation (B) The maximum kinetic energy of the electrons emitted depends on the intensity of the incident radiation (C) The maximum kinetic energy of the electrons emitted depends on the frequency of the incident radiation (D) The value of the stopping potential increases with the increase in frequency of the incident radiation.

   ›Reveal solutionSolution

   The photoelectric effect shows that the maximum kinetic energy of emitted electrons depends on frequency, not intensity — so statement (B) is wrong.

   The photoelectric effect is a cornerstone of quantum physics. It demonstrates that light behaves as discrete packets of energy (photons), and the energy of each photon is proportional to its frequency, not its amplitude (intensity). This is why intensity only affects the number of electrons emitted, not their individual energies. Let’s examine each statement carefully.

   1. Statement (A): Photo electric current is directly proportional to the intensity of the incident radiation.

      This is correct. Intensity means the number of photons per second. More photons → more electrons ejected → larger current (provided the frequency is above the threshold). So (A) is true.

   2. Statement (B): The maximum kinetic energy of the electrons emitted depends on the intensity of the incident radiation.

      This is the classic pitfall. In the photoelectric effect, each electron absorbs one photon. The photon’s energy is $hf$. After overcoming the work function $\phi$, the leftover energy becomes kinetic energy:

   $K_{\text{max}} = hf - \phi$

   Intensity changes the number of photons, not the energy per photon. So $K_{\text{max}}$ is independent of intensity. Thus (B) is false.

   3. Statement (C): The maximum kinetic energy of the electrons emitted depends on the frequency of the incident radiation. This is correct, as shown by the equation above. Higher frequency → larger $K_{\text{max}}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. COMEDK 2025Set 2025-E1 markMCQ
   Q.The stopping potential when a metal surface is illuminated by light of wavelength $\lambda$ is 15 V . The stopping potential when the same surface is illuminated by light of wavelength $4 \lambda$ is 3 V . The ratio of threshold wavelength to the initial incident wavelength $\lambda$ is: (A) $1: 16$ (B) $1: 15$ (C) $16: 1$ (D) $15: 1$

   ›Reveal solutionSolution

   Using the photoelectric equation for two different wavelengths, we eliminate the work function to find the threshold wavelength; the ratio of threshold wavelength to $\lambda$ is $16:1$, so the correct option is (C).

   Concept & Intuition

   The photoelectric effect tells us that the maximum kinetic energy of ejected electrons is $K_{\text{max}} = hf - \phi$, where $\phi$ is the work function. The stopping potential $V_s$ is related by $eV_s = K_{\text{max}}$. Since $f = c/\lambda$, we can write $eV_s = \frac{hc}{\lambda} - \phi$. With two data points, we can solve for both $\phi$ and $hc$, then find the threshold wavelength $\lambda_{\text{th}} = hc/\phi$. The ratio $\lambda_{\text{th}} / \lambda$ follows directly.

   Step-by-step solution

   1. Write the photoelectric equation for each case For wavelength $\lambda$, stopping potential $V_1 = 15$ V:

   $e \cdot 15 = \frac{hc}{\lambda} - \phi \quad \text{(1)}$

   For wavelength $4\lambda$, stopping potential $V_2 = 3$ V:

   $e \cdot 3 = \frac{hc}{4\lambda} - \phi \quad \text{(2)}$

   2. Subtract equation (2) from equation (1) to eliminate $\phi$

   $15e - 3e = \frac{hc}{\lambda} - \frac{hc}{4\lambda}$

   $12e = \frac{hc}{\lambda} \left(1 - \frac{1}{4}\right) = \frac{hc}{\lambda} \cdot \frac{3}{4}$

   So

   $\frac{hc}{\lambda} = 12e \cdot \frac{4}{3} = 16e$

   This tells us the photon energy for wavelength $\lambda$ is $16$ eV (since $e$ is the electron charge, $16e$ means 16 eV).

   3. Find the work function $\phi$ Substitute $\frac{hc}{\lambda} = 16e$ into equation (1):

   $15e = 16e - \phi \quad \Rightarrow \quad \phi = e$

   So the work function is $1$ eV.

   4. Determine the threshold wavelength …

   🔒 Start your 14-day free trial to see the full solution →
7. COMEDK 2025Set 2025-M1 markMCQ
   Q.Emission of electrons from a metal plate illuminated with monochromatic electromagnetic radiation will always take place provided (A) The plate is positively charged (B) The plate is negatively charged (C) The radiation is sufficiently intense (D) The work function of the plate is less than the energy of a single photon and the plate is uncharged

   ›Reveal solutionSolution

   The key idea is the photoelectric effect: emission occurs when a single photon’s energy exceeds the work function, regardless of intensity or charge, so the correct condition is that the photon energy is greater than the work function and the plate is uncharged (to avoid retarding fields).

   The photoelectric effect is a quantum phenomenon: light behaves as particles (photons), each carrying energy $E = hf$, where $h$ is Planck’s constant and $f$ is the frequency. An electron in the metal needs a minimum energy—the work function $\phi$—to escape. If a single photon’s energy is less than $\phi$, no electron can be ejected, no matter how many photons (i.e., how intense the light) you throw at it. Intensity only affects the number of emitted electrons, not whether emission starts. The plate’s charge matters because a positive charge attracts electrons (helping emission) and a negative charge repels them (hindering it), but the necessary condition is purely energetic.

   Let’s examine each option step by step.

   1. Option (A): The plate is positively charged

      A positive charge lowers the barrier for electrons to leave, but if the photon energy is below the work function, no electron can absorb enough energy to escape. Positive charge alone cannot create emission; it only assists if emission is already possible. So this is not sufficient.

   2. Option (B): The plate is negatively charged

      A negative charge repels electrons, making it harder for them to leave. Even if the photon energy exceeds the work function, a strong negative charge can prevent emission. This is not a condition that ensures emission; it often prevents it.

   3. Option (C): The radiation is sufficiently intense

      Intensity means more photons per second, but each photon still carries the same energy $hf$. If $hf < \phi$, no single photon can eject an electron—classical wave theory fails here. Increasing intensity only increases the number of photons, not their individual energy. So this is incorrect.

   4. Option (D): The work function of the plate is less than the energy of a single photon and the plate is uncharged …

   🔒 Start your 14-day free trial to see the full solution →
8. KCET 2024Set D-21 markMCQ
   Q.The photoelectric work function for photo metal is $2.4$ eV. Among the four wavelengths, the wavelength of light for which photo-emission does not take place is (A) $200$ nm (B) $300$ nm (C) $700$ nm (D) $400$ nm

   ›Reveal solutionSolution

   Photo-emission requires photon energy $\ge$ work function. The threshold wavelength is $\lambda_0 = \frac{hc}{\phi} \approx 517$ nm. Light with wavelength longer than this (lower energy) cannot cause emission. Among the options, 700 nm is the only one above threshold, so no photo-emission occurs for that.

   The key idea is the photoelectric effect: an electron is ejected from a metal surface only if the incoming photon carries enough energy to overcome the binding energy (the work function $\phi$). If the photon’s energy is less than $\phi$, no electron is emitted, no matter how intense the light.

   The energy of a photon is $E = \frac{hc}{\lambda}$, where $h$ is Planck’s constant, $c$ is the speed of light, and $\lambda$ is the wavelength. For photo-emission to occur, we need $E \ge \phi$. So the condition for no emission is simply $E < \phi$, i.e. $\lambda > \frac{hc}{\phi}$.

   Let’s find that threshold wavelength.

   1. Write the work function in joules.

      $\phi = 2.4$ eV. Since $1$ eV $= 1.6 \times 10^{-19}$ J,

      $\phi = 2.4 \times 1.6 \times 10^{-19} = 3.84 \times 10^{-19}$ J.

   2. Use the photon energy formula.

      $hc = 6.63 \times 10^{-34} \times 3 \times 10^8 = 1.989 \times 10^{-25}$ J·m.

      The threshold wavelength $\lambda_0$ is where $hc/\lambda_0 = \phi$, so

      $\lambda_0 = \frac{hc}{\phi} = \frac{1.989 \times 10^{-25}}{3.84 \times 10^{-19}} \approx 5.18 \times 10^{-7}$ m $= 518$ nm.

      Tip

      A faster route: use $hc = 1240$ eV·nm (a standard constant). Then $\lambda_0 = \frac{1240}{2.4} \approx 517$ nm. This avoids unit conversions entirely. …

   🔒 Start your 14-day free trial to see the full solution →
9. COMEDK 2024Set 2024-E1 markMCQ
   Q.A photon emitted during the de-excitation of electron from a state $\mathrm{n}$ to the second excited state in a hydrogen atom, irradiates a metallic electrode of work function $0.5 \mathrm{~eV}$, in a photocell, with a stopping voltage of $0.47 \mathrm{~V}$. Obtain the value of quantum number of the state '$n$'. (A) 5 (B) 6 (C) 4 (D) 3

   ›Reveal solutionSolution

   The problem combines the Bohr model for hydrogen with the photoelectric effect: the photon energy from the transition $n \to 3$ equals the work function plus the stopping voltage energy, giving $n = 5$. The correct option is (A).

   We are told the electron de‑excites from some state $n$ to the second excited state. In hydrogen, the ground state is $n=1$, first excited state is $n=2$, and second excited state is $n=3$. So the transition is $n \to 3$.

   The emitted photon then hits a metal surface (work function $\phi = 0.5\ \text{eV}$) and ejects electrons. The stopping voltage $V_s = 0.47\ \text{V}$ tells us the maximum kinetic energy of the photoelectrons:

   $K_{\text{max}} = e V_s = 0.47\ \text{eV}.$

   By the photoelectric equation:

   $E_{\text{photon}} = \phi + K_{\text{max}} = 0.5\ \text{eV} + 0.47\ \text{eV} = 0.97\ \text{eV}.$

   Now we need the energy of a photon emitted when an electron in hydrogen drops from level $n$ to level $3$. The energy levels in hydrogen are:

   $E_n = -\frac{13.6\ \text{eV}}{n^2}.$

   So the photon energy is:

   $E_{\text{photon}} = E_n - E_3 = 13.6\left(\frac{1}{3^2} - \frac{1}{n^2}\right)\ \text{eV}.$

   Set this equal to $0.97\ \text{eV}$:

   $13.6\left(\frac{1}{9} - \frac{1}{n^2}\right) = 0.97.$

   1. Divide both sides by 13.6:

   $\frac{1}{9} - \frac{1}{n^2} = \frac{0.97}{13.6} \approx 0.07132.$

   2. Then:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
10. COMEDK 2024Set 2024-M1 markMCQ
    Q.The threshold frequency for a metal surface is '$n_0$'. A photo electric current '$I$' is produced when it is exposed to a light of frequency $\left(\frac{11}{6}\right) \mathrm{n}_{\mathrm{o}}$ and intensity $\mathrm{I}_{\mathrm{n}}$. If both the frequency and intensity are halved, the new photoelectric current '$\mathrm{I}^1$' will become: (A) $\mathrm{I}^1=\frac{1}{4} \mathrm{I}$ (B) $\mathrm{I}^1=2 \mathrm{I}$ (C) $\mathrm{I^1=0}$ (D) $\mathrm{I}^1=\frac{1}{2} \mathrm{I}$

    ›Reveal solutionSolution

    The photoelectric current depends on the number of photoelectrons emitted per second, which is proportional to the intensity of light only if the photon energy exceeds the work function. Halving the frequency from $\frac{11}{6}n_0$ to $\frac{11}{12}n_0$ makes it below the threshold $n_0$, so no photoelectrons are emitted — the new current is zero. The correct option is (C).

    ---

    Concept and Intuition

    The photoelectric effect has a sharp threshold: no electrons are emitted if the light frequency is below the threshold frequency $n_0$, no matter how intense the light. The current is proportional to the number of photoelectrons per second, which equals the number of photons per second only if each photon has enough energy ($h f \ge h n_0$) to kick out an electron. If the frequency drops below $n_0$, even a flood of photons produces zero current.

    Here, the initial frequency is $\frac{11}{6}n_0 \approx 1.833 n_0$ — safely above threshold. Halving it gives $\frac{11}{12}n_0 \approx 0.917 n_0$, which is below threshold. So the current must become zero, regardless of intensity changes.

    ---

    Step-by-step reasoning

    1. Recall the condition for photoelectric emission

       A photon of frequency $f$ has energy $E = h f$. It can eject an electron only if $h f \ge h n_0$, i.e. $f \ge n_0$. Below $n_0$, no electrons are emitted — the current is zero.

    2. Initial situation

       Frequency: $f_1 = \frac{11}{6} n_0 > n_0$ → emission occurs.

       Intensity: $I_n$ (some value). The photoelectric current $I$ is proportional to the number of photons per second (since each photon above threshold can release one electron, assuming no other losses). So $I \propto \text{(intensity)}$.

    3. New situation — frequency halved

       New frequency:

    $f_2 = \frac{1}{2} \times \frac{11}{6} n_0 = \frac{11}{12} n_0$

    Compare with threshold:

    $\frac{11}{12} n_0 < n_0$

    Hence $f_2 < n_0$. No photoelectrons are emitted — the current must be zero.

    4. What about the intensity being halved? …

    🔒 Start your 14-day free trial to see the full solution →
11. KCET 2023Set A-31 markMCQ
    Q.In the following equation representing $\beta^-$ decay, the number of neutrons in the nucleus X is ${}_{83}^{210}\text{Bi} \rightarrow X + e^{-1} + \bar{\nu}$ (A) 127 (B) 125 (C) 84 (D) 126

    ›Reveal solutionSolution

    In $\beta^-$ decay, a neutron converts into a proton, so the mass number stays the same but the atomic number increases by 1. The daughter nucleus X has 126 neutrons.

    The concept: $\beta^-$ decay conserves mass number but changes atomic number.

    In $\beta^-$ decay, a neutron inside the nucleus transforms into a proton, an electron, and an antineutrino. The electron and antineutrino are ejected. The key point: the mass number $A$ (total protons + neutrons) does not change, because a neutron is simply replaced by a proton. The atomic number $Z$ (number of protons) increases by 1, since a new proton appears.

    So for the reaction ${}_{83}^{210}\text{Bi} \rightarrow X + e^{-1} + \bar{\nu}$, the parent bismuth has $Z = 83$ and $A = 210$. After decay, the daughter nucleus X must have the same $A = 210$, but $Z = 83 + 1 = 84$.

    Now, the number of neutrons in X is $A - Z = 210 - 84 = 126$.

    Let’s walk through it step by step.

    1. Identify the parent nucleus. Bismuth-210: atomic number $Z = 83$, mass number $A = 210$. So it has $83$ protons and $210 - 83 = 127$ neutrons.

    2. Apply the $\beta^-$ decay rule. A neutron changes into a proton. So the total number of nucleons stays $210$, but the proton count goes up by one. The daughter nucleus X therefore has $Z = 84$ and $A = 210$. …

    🔒 Start your 14-day free trial to see the full solution →
12. COMEDK 2023Set 2023-E1 markMCQ
    Q.In the photoelectric experiment, the frequency of the incident radiation is doubled. What will be its effect on the photoelectric current? (A) Photoelectric current will be halved (B) Photoelectric current will be doubled (C) Photoelectric current will not change. (D) Photoelectric current will become zero

    ›Reveal solutionSolution

    Photoelectric current is governed by intensity (photon rate), not frequency, so doubling the frequency does not change it.

    In the photoelectric effect, the saturation photoelectric current is proportional to the number of photoelectrons emitted per second, which in turn is proportional to the number of incident photons per second — i.e. the intensity of the light. …

    🔒 Start your 14-day free trial to see the full solution →