Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successiv

I Multiple Choice Questions · Q8

Q. Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of the emitted electrons will be

(a) 1 : 4

(b) 1 : 3

(d) 1 : 9

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Concept understanding — Maximum Kinetic Energy

Maximum Kinetic Energy – From Intuition to Precision

Kinetic energy is the energy of motion: the faster something moves, the more kinetic energy it has. In many physical situations there is a maximum possible kinetic energy a particle can reach — set either by energy conservation or by an external energy constraint.

The Precise Statement

K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2

Where:

$K_{\text{max}}$ = maximum kinetic energy (in joules)
$m$ = mass of the object (in kg)
$v_{\text{max}}$ = maximum speed reached (in m/s)

This formula alone doesn't tell you why there's a maximum — the physics lies in energy conservation or in an external constraint that limits the speed.

Where Does the Maximum Come From?

1. Energy conservation (no friction)

In a closed system, total mechanical energy $E = K + U$ is constant, so

K_{\text{max}} = E - U_{\text{min}}

The maximum kinetic energy occurs when the potential energy $U$ is at its minimum — for example, a falling object is fastest (and $U$ smallest) just before it lands.

2. External constraints (e.g., the photoelectric effect)

In modern physics, electrons in a metal absorb light energy. Each photon delivers a fixed energy $hf$ . The electron must spend part of that energy escaping the metal (the work function $\phi$ ); the rest becomes kinetic energy:

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

Here the maximum is set entirely by the photon energy — no matter how intense the light, no single electron can gain more kinetic energy than this.

A Common Mistake

Watch out

Students often think "maximum kinetic energy" means the fastest speed possible in the universe. It doesn't. The "maximum" is relative to the given system — the highest value under the stated conditions (height, spring compression, photon energy, and so on), not a universal speed limit. …

Why this formula?

Maximum Kinetic Energy — Why the Formula Holds

The idea of "maximum kinetic energy" appears in two very different contexts in your syllabus: photoelectric effect (modern physics) and simple harmonic motion (oscillations). I'll cover both, because the why is different in each case.

1. In the Photoelectric Effect

The formula you must know:

K_{\text{max}} = h\nu - \phi

where $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\phi$ is the work function of the metal.

Why this formula? It comes from Einstein's photon model and energy conservation.

A single photon carries energy $E = h\nu$ . When it strikes a metal surface, it can transfer all of its energy to one electron. That electron must first overcome the binding force holding it in the metal — the minimum energy needed for this is the work function $\phi$ . Any leftover energy becomes the electron's kinetic energy after it escapes.

So:

Photon energy = Energy to escape + Kinetic energy of ejected electron

h\nu = \phi + K

If the electron just barely escapes (with zero kinetic energy), the photon frequency is the threshold frequency $\nu_0$ , where $h\nu_0 = \phi$ .

For a higher frequency, the maximum kinetic energy an ejected electron can have is when it absorbs the photon's full energy and loses nothing to collisions inside the metal. That gives:

K_{\text{max}} = h\nu - \phi

Watch out

$K_{\text{max}}$ does not depend on light intensity. Intensity only increases the number of electrons, not their maximum energy. This was the key puzzle that classical physics couldn't explain.

2. In Simple Harmonic Motion (SHM)

For a particle executing SHM, the maximum kinetic energy is:

K_{\text{max}} = \frac{1}{2} m \omega^2 A^2

where $m$ is mass, $\omega$ is angular frequency, and $A$ is amplitude.

Why this formula? It follows directly from the velocity equation.

In SHM, displacement is $x = A \sin(\omega t + \phi)$ . Differentiating gives velocity:

v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)

The velocity is maximum when $\cos(\omega t + \phi) = \pm 1$ , i.e., at the equilibrium position ( $x = 0$ ):

$v_{\text{max}} = A\omega$ …

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Step 1. Using $K_{max}=E_{photon}-\phi_0$ : for the 0.9 eV photon, $K_1=0.9-0.6=0.3$ eV; for the 3.3 eV photon, $K_2=3.3-0.6=2.7$ eV.

Step 2. Since $K=\tfrac12mv^2$ , speed scales as $v\propto\sqrt K$ , so $\dfrac{v_1}{v_2}=\sqrt{\dfrac{K_1}{K_2}}=\sqrt{\dfrac{0.3}{2.7}}=\sqrt{\dfrac19}=\dfrac13$ .

Step 3. So the ratio of maximum speeds is $v_1:v_2=1:3$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

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), …

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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: …

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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:

…

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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. T …

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CBSE 2025Set ANNUAL1 markMCQ
$= \dfrac{1}{2}mv^2$
(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.

…

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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 $\phi_0$ .

…

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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:

…

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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 o …

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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 alread …

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

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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:

…

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

$KE_1 = 1 - 0.5 = 0.5\text{ eV}$ …

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Q.Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of the emitted electrons will be

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