In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15}\ \text{V s}$. Calculate the value of Planck's constant.

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light

Exercises · 11.6

4.12 \times 10^{-15}\ \text{V s}

Punjab PsebTextbookSubjective· 2mImportance★★★★★est

11% · 9/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

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

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 ( $x = 0$ ):

$v_{\text{max}} = A\omega$ …

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

The slope of the $V_0$ vs $\nu$ graph equals $h/e$ . Using the given slope $4.12 \times 10^{-15}\ \text{V s}$ and $e = 1.6 \times 10^{-19}\ \text{C}$ , Planck's constant is $h = 6.592 \times 10^{-34}\ \text{J s}$ .

The photoelectric effect experiment gives us a direct, clean way to measure Planck's constant. When you shine light of different frequencies on a metal surface and measure the stopping potential (the voltage that just stops the most energetic electrons), you get a straight line when you plot cut-off voltage $V_0$ against frequency $\nu$ .

Why is this line so important? Because Einstein's photoelectric equation tells us:

eV_0 = h\nu - \phi

Here $e$ is the electron charge, $h$ is Planck's constant, and $\phi$ is the work function of the metal. Rearranging:

V_0 = \frac{h}{e}\nu - \frac{\phi}{e}

This is the equation of a straight line $y = mx + c$ , where $y = V_0$ , $x = \nu$ , the slope $m = h/e$ , and the intercept $c = -\phi/e$ .

So the slope of the graph directly gives $h/e$ . That's the key insight — we don't need to know the work function or any other property of the metal. The slope alone is enough.

Identify what the slope represents. From the equation $V_0 = (h/e)\nu - \phi/e$ , the slope $m = h/e$ . The problem gives $m = 4.12 \times 10^{-15}\ \text{V s}$ .
Recall the electron charge. The elementary charge $e = 1.6 \times 10^{-19}\ \text{C}$ (this is a standard value you must remember for such problems).
Solve for $h$ . Since $m = h/e$ , we have $h = m \times e$ .

h = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19})

Multiply the numbers: $4.12 \times 1.6 = 6.592$ …

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

Students often lose marks on this exact problem not because the calculation is hard, but because they miss the physical link between the slope and Planck's constant. Let's go through the common mistakes one by one.

Mistake 1: Forgetting that the slope is $h/e$ , not $h$

The cut-off voltage $V_0$ is related to frequency $f$ by:

eV_0 = hf - \phi

So:

V_0 = \frac{h}{e} f - \frac{\phi}{e}

The slope of the $V_0$ vs $f$ graph is $\dfrac{h}{e}$ , not $h$ itself.

Watch out

Many students directly write $h = \text{slope}$ and get $h = 4.12 \times 10^{-15}\ \text{J s}$ , which is wrong by a factor of about $1.6 \times 10^{-19}$ .

How to avoid: Always write the equation $eV_0 = hf - \phi$ first. Then rearrange to $V_0 = (h/e)f - \phi/e$ . The slope is $h/e$ . Never skip this step.

Mistake 2: Using the wrong unit for Planck's constant

The slope is given in $\text{V s}$ (volt-seconds). Planck's constant must be in $\text{J s}$ (joule-seconds). Since $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ , you must multiply by $e$ to convert.

Note

$h = \text{slope} \times e$

$h = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19})$

$h = 6.592 \times 10^{-34}\ \text{J s}$

How to avoid: Always check the unit of the given quantity. If it's in $\text{V s}$ , you need to multiply by $e$ (in coulombs) to get $\text{J s}$ . If the slope were given in $\text{eV s}$ , then $h$ would equal the slope directly.

Mistake 3: Using $e = 1.6 \times 10^{-19}\ \text{C}$ incorrectly or forgetting the sign

The electronic charge $e$ is $1.6 \times 10^{-19}\ \text{C}$ (positive magnitude). Some students use $e = 1.6 \times 10^{-19}\ \text{J/eV}$ when converting, but here you need the charge in coulombs because $V_0$ is in volts and $eV_0$ gives energy in joules.

Tip

Remember: $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ is an energy conversion. But $e = 1.6 \times 10^{-19}\ \text{C}$ is a charge. In $eV_0$ , $e$ is charge, $V_0$ is potential difference, product is energy in joules.

How to avoid: Write the units explicitly in your working:

\text{slope} = 4.12 \times 10^{-15}\ \frac{\text{V}}{\text{Hz}} = 4.12 \times 10^{-15}\ \text{V s}

Then:

h = (\text{slope}) \times e = (4.12 \times 10^{-15}\ \text{V s}) \times (1.6 \times 10^{-19}\ \text{C})

Since $1\ \text{V} \times 1\ \text{C} = 1\ \text{J}$ , you get $h$ in $\text{J s}$ .

Mistake 4: Rounding off too early …

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

←Previous question11.5 — The energy flux of sunlight reaching the surface of the earth is 1.388…→Next question11.7 — A 100W sodium lamp radiates energy uniformly in all directions. The la…

Q.In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15}\ \text{V s}$ . Calculate the value of Planck's constant.

Concept understanding — Maximum Kinetic Energy

Maximum Kinetic Energy – From Intuition to Precision

The Precise Statement

Where Does the Maximum Come From?

A Common Mistake

Why this formula?

Maximum Kinetic Energy — Why the Formula Holds

1. In the Photoelectric Effect

2. In Simple Harmonic Motion (SHM)

Method: Slope of Cut-Off Voltage vs Frequency

Why this works

Steps

Mistake 1: Forgetting that the slope is $h/e$ , not $h$

Mistake 2: Using the wrong unit for Planck's constant

Mistake 3: Using $e = 1.6 \times 10^{-19}\ \text{C}$ incorrectly or forgetting the sign

Mistake 4: Rounding off too early …

Q.In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×10−15 V s4.12 \times 10^{-15}\ \text{V s}4.12×10−15 V s. Calculate the value of Planck's constant.

🎓Unlock everything free for 14 days

Q.In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15}\ \text{V s}$ . Calculate the value of Planck's constant.

Unlock everything free for 14 days