Q.Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for a photoelectric emission experiment.

Showing the 12 most recent of 19 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.The maximum kinetic energy of the photoelectrons varies (A) linearly with the frequency and the intensity of the incident radiation (B) linearly with the frequency and is independent of the intensity of the incident radiation (C) inversely with the frequency and is independent of the intensity of the incident radiation (D) inversely with the intensity and is independent of the frequency of the incident radiation

   ›Reveal solutionSolution

   Einstein's photoelectric equation makes Kₘₐₓ a linear function of frequency, independent of intensity.

   Einstein's photoelectric equation is:

   $K_{max}=h\nu-\phi_0$

   where $\nu$ is the frequency of incident light and $\phi_0$ the work function. This is a straight line in $\nu$ (slope $h$), so $K_{max}$ varies linearly with frequency. Intensity only sets the number of photons (hence the photocurrent), not the energy per photon, so $K_{max}$ is independent of intensity.

   ✓Final answer

   (B) linearly with the frequency and is independent of the intensity of the incident radiation.

2. CBSE 2026Set A1 markMCQ
   Q.The maximum velocity of an electron emitted from a metal surface becomes two times when the frequency v of the incident light is doubled. The work function of the metal is (A) 2hv/3 (B) hv/3 (C) zero (D) hv/2

   ›Reveal solutionSolution

   Using KE ∝ v_max², doubling v_max means KE becomes 4×; solving Einstein's equation at frequency v and 2v gives φ = 2hv/3.

   Einstein's photoelectric equation is

   $\tfrac{1}{2}m v_{max}^2 = h\nu - \phi.$

   At frequency $\nu$, maximum speed $v_{max}$:

   $\tfrac{1}{2}m v_{max}^2 = h\nu - \phi. \quad (1)$

   When the frequency is doubled to $2\nu$, the maximum speed doubles to $2v_{max}$, so the kinetic energy becomes four times:

   $\tfrac{1}{2}m (2v_{max})^2 = 4\left(\tfrac{1}{2}m v_{max}^2\right) = h(2\nu) - \phi. \quad (2)$

   Substitute (1) into (2):

   $4(h\nu - \phi) = 2h\nu - \phi$

   $4h\nu - 4\phi = 2h\nu - \phi$

   $2h\nu = 3\phi \;\Rightarrow\; \phi = \frac{2h\nu}{3}.$

   ✓Final answer

   (A) 2hv/3 — the work function of the metal is φ = 2hν/3.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The specific charge of electron is :

   (a) 1.9 × 10^31 C/kg

   (b) 1.6 × 10^19 C/kg

   (c) 1.76 × 10^11 C/kg

   (d) 1.76 × 10^-11 C/kg

   ›Reveal solutionSolution

   e/m = (1.6 × 10⁻¹⁹)/(9.1 × 10⁻³¹) ≈ 1.76 × 10¹¹ C/kg.

   The specific charge of the electron is the ratio of its charge to its mass, e/m. Using e = 1.6 × 10⁻¹⁹ C and m = 9.1 × 10⁻³¹ kg:

   e/m = (1.6 × 10⁻¹⁹) / (9.1 × 10⁻³¹) = 1.76 × 10¹¹ C/kg.

   This value was first measured by J. J. Thomson.

   ✓Final answer

   (c) 1.76 × 10¹¹ C/kg.

4. CBSE 2026Set ANNUAL1 mark
   Q.Write True or False: By suitably heating, sufficient thermal energy can be imparted to the free electrons to enable them to come out of the metal.

   ›Reveal solutionSolution

   True — this describes thermionic emission.

   Free electrons in a metal are normally held inside by the surface potential barrier (the work function). If the metal is heated to a high temperature, the free electrons gain thermal energy; some acquire enough energy to overcome the work function and are emitted from the metal surface. This process is called thermionic emission (used in thermionic valves and electron guns).

   Hence the given statement is True.

   ✓Final answer

   True.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Assertion (A): When ultraviolet light is incident on two photosensitive metals having different work functions then maximum kinetic energy of the photo electrons is greater for metal having low work function. Reason (R): Kinetic energy $= \dfrac{1}{2}mv^2$

   (a) Both A and R are correct and R is the correct explanation of A.

   (b) Both A and R are correct but R is not correct explanation of A.

   (c) A is correct but R is incorrect.

   (d) A and R both are incorrect.

   ›Reveal solutionSolution

   A is true by Einstein's photoelectric equation; R is a true but generic definition of KE that, by itself, does not explain why a lower work function gives a higher maximum photoelectron KE.

   Assertion (A) is true. By Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is

   $KE_{max} = h\nu - \phi$

   where $\phi$ is the work function of the metal. For the same incident frequency $\nu$, a metal with a lower work function $\phi$ gives a larger $KE_{max}$, so A is correct.

   Reason (R), $KE=\tfrac{1}{2}mv^2$, is also a true statement — it is simply the general classical definition of kinetic energy. However, this generic definition does not by itself explain why the work function affects the photoelectrons' maximum kinetic energy; the actual explanation is Einstein's photoelectric equation ($KE_{max}=h\nu-\phi$), which involves the photon energy and the work function, not the bare formula $\tfrac12 mv^2$. So R, though true, is not the correct explanation of A.

   ✓Final answer

   (b) Both A and R are correct but R is not correct explanation of A.

6. CBSE 2025Set IMPROVEMENT1 markMCQ
   Q.Assertion (A): Photoelectric current is zero when the stopping potential (V₀) is sufficient to repel even the most energetic (K_max) photoelectrons. Reason (R): According to Einstein's photoelectric equation, K_max = hν − ϕ₀, where hν is the energy of each quantum of radiation incident on the metal surface and ϕ₀ is the work function of the metal. Select the correct option.

   (a) Both A and R are correct and R is the correct explanation of A.

   (b) Both A and R are correct but R is not the correct explanation of A.

   (c) A is correct but R is incorrect.

   (d) Both A and R are incorrect.

   ›Reveal solutionSolution

   Both statements are individually true, but Einstein's equation for K_max does not itself explain why the photoelectric current becomes zero at the stopping potential.

   Assertion (A) is true by the very definition of the stopping potential $V_0$: it is the minimum retarding potential that is just able to stop even the fastest (most energetic, $K_{max}$) photoelectrons, so at $V = V_0$ no photoelectron — however energetic — can reach the collector, and the photoelectric current drops to zero.

   Reason (R) states Einstein's photoelectric equation, $K_{max} = h\nu - \phi_0$, which is also true — it correctly gives the maximum kinetic energy of an emitted photoelectron in terms of the incident photon energy $h\nu$ and the work function $\phi_0$.

   However, R by itself does not explain WHY the current becomes zero at $V_0$ — that fact instead follows from the definition/role of the stopping potential (namely, $eV_0 = K_{max}$, i.e. the work done against the retarding field equals $K_{max}$), not directly from the formula for $K_{max}$ in terms of $h\nu$ and $\phi_0$. So R, though true, is not the correct explanation of A.

   ✓Final answer

   (b) Both A and R are correct but R is not the correct explanation of A.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The energy of a photon is 18 eV and the work function of the material is 8 eV. The value of stopping potential is

   (a) zero

   (b) 8 V

   (c) 10 V

   (d) 26 V

   ›Reveal solutionSolution

   The stopping potential (in volts) equals the maximum kinetic energy of the photoelectrons (in eV), which is the photon energy minus the work function.

   Einstein's photoelectric equation:

   $K_{max} = h\nu - \phi_0$

   where $h\nu$ is the photon energy and $\phi_0$ is the work function. Given $h\nu = 18$ eV and $\phi_0 = 8$ eV:

   $K_{max} = 18 - 8 = 10\ \text{eV}$

   The stopping potential $V_0$ is defined by $eV_0 = K_{max}$, so numerically (using eV as the energy unit) $V_0 = 10$ V.

   ✓Final answer

   (c) 10 V.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The maximum kinetic energy of the photoelectrons depends only on :

   (a) incident angle.

   (b) frequency.

   (c) pressure.

   (d) potential.

   ›Reveal solutionSolution

   By Einstein's photoelectric equation, $KE_{max} = h\nu - \phi_0$, so for a given metal (fixed work function $\phi_0$) the maximum KE depends only on the frequency $\nu$ of incident light, not on its intensity.

   Einstein's photoelectric equation is

   $KE_{max} = h\nu - \phi_0$

   where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\phi_0$ is the work function of the metal (a fixed property of the material). Increasing the intensity of light increases the number of photoelectrons emitted (photocurrent) but not the maximum kinetic energy of each electron, since each photon still carries energy $h\nu$. Angle of incidence and pressure play no role in this energy relation.

   ✓Final answer

   Maximum kinetic energy of photoelectrons depends only on the frequency of incident light — option (b).

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.The maximum kinetic energy of photoelectrons emitted from a metal surface when photons of energy 5.6 eV fall on it, is 4eV. The stopping potential in volts is –

   (a) 1.6 V

   (b) 3.2V

   (c) 4V

   (d) 5.6V

   ›Reveal solutionSolution

   Stopping potential $V_0 = KE_{max}/e$, and $KE_{max}$ is already given directly as 4 eV.

   The stopping potential is defined by $eV_0 = KE_{max}$. Here $KE_{max}$ is given directly as $4\,$eV (not needing to be computed from the photon energy and work function, both of which are already reflected in the given value):

   $V_0 = \frac{KE_{max}}{e} = \frac{4\,\text{eV}}{e} = 4\,\text{V}$

   ✓Final answer

   (c) 4V.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then, the maximum possible velocity of the emitted electron will be :

    (a) $2\sqrt{\dfrac{h\nu_0}{m}}$

    (b) $\sqrt{\dfrac{h\nu_0}{m}}$

    (c) $\sqrt{\dfrac{h\nu_0}{2m}}$

    (d) $\sqrt{\dfrac{6h\nu_0}{m}}$

    ›Reveal solutionSolution

    With incident frequency $4\nu_0$, Einstein's photoelectric equation gives $KE_{max}=3h\nu_0$, so $v_{max}=\sqrt{6h\nu_0/m}$.

    Working

    Einstein's photoelectric equation: $KE_{max} = h\nu - h\nu_0$.

    Given $\nu = 4\nu_0$:

    $KE_{max} = h(4 u_0) - h u_0 = 3h u_0$

    Since $KE_{max} = \dfrac12 m v_{max}^2$:

    $\dfrac12mv_{max}^2 = 3h u_0$

    $v_{max} = \sqrt{\dfrac{6h u_0}{m}}$

    ✓Final answer

    The correct option is (d): $v_{max}=\sqrt{\dfrac{6h\nu_0}{m}}$

11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The maximum kinetic energy of a photo electron emitted from a metal is 1.8 eV. The value of stopping potential (cut-off voltage) will be -

    (a) 3.6 V

    (b) 2.0 V

    (c) 1.8 V

    (d) 0.9 V

    ›Reveal solutionSolution

    The stopping potential V₀ is defined by eV₀ = KE_max, so measuring KE_max in electron-volts directly gives V₀ in volts.

    The stopping potential $V_0$ is the retarding potential just sufficient to stop even the fastest photoelectrons, defined by:

    $eV_0 = KE_{max}$

    Here $KE_{max} = 1.8\ \text{eV}$, i.e. $KE_{max} = 1.8\,e$ (in joules, using the electron-volt definition). Substituting:

    $eV_0 = 1.8\,e \implies V_0 = 1.8\ \text{V}$

    This is exactly why the electron-volt is a convenient unit — the numeric value of KE in eV equals the stopping potential in volts.

    ✓Final answer

    (c) 1.8 V.

12. CBSE 2024Set ANNUAL1 markMCQ
    Q.Photons of energies 1eV and 2eV are successively incident on a metallic surface of work function 0.5 eV. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be -

    (a) 1:2

    (b) 1:3

    (c) 1:1

    (d) 1:4

    ›Reveal solutionSolution

    Einstein's photoelectric equation gives $KE_{max} = E_{photon} - \phi$; subtract the same work function from each photon energy and take the ratio.

    Work function $\phi = 0.5\text{ eV}$.

    $KE_1 = 1 - 0.5 = 0.5\text{ eV}$

    $KE_2 = 2 - 0.5 = 1.5\text{ eV}$

    $\dfrac{KE_1}{KE_2} = \dfrac{0.5}{1.5} = \dfrac{1}{3}$

    ✓Final answer

    $KE_1:KE_2 = 1:3$ — option (b).