Consider a metal exposed to light of wavelength 600nm. The maximum energy of the electron doubles when light o

NCERT Exemplar · Q14

600\ \text{nm}

Mahe DhseShort· 3mImportance★★★★★est

65% · 54/83 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 problem uses the photoelectric equation $K_{\text{max}} = hf - \phi$ for two wavelengths. The ratio of maximum kinetic energies is given (doubles), which lets us eliminate the unknown $K_{\text{max}}$ and solve directly for the work function $\phi$ . The answer is $\phi \approx 1.03\ \text{eV}$ .

The photoelectric effect tells us that when light hits a metal, an electron can be ejected if the photon energy exceeds the work function. The leftover energy becomes the electron’s maximum kinetic energy. Here, we have two different wavelengths, and we know that the maximum kinetic energy for the second case is exactly twice that of the first. That’s the only link between the two situations — and it’s enough to find the work function.

Let’s set it up.

Write the photoelectric equation for each case. For wavelength $\lambda_1 = 600\ \text{nm}$ , let the maximum kinetic energy be $K_1$ . For $\lambda_2 = 400\ \text{nm}$ , it’s $K_2$ , and we’re told $K_2 = 2K_1$ . The photon energy is $hc/\lambda$ , so:

K_1 = \frac{hc}{\lambda_1} - \phi

K_2 = \frac{hc}{\lambda_2} - \phi

Use the doubling condition. Since $K_2 = 2K_1$ , substitute:

\frac{hc}{\lambda_2} - \phi = 2\left(\frac{hc}{\lambda_1} - \phi\right)

Solve for $\phi$ . Expand the right side:

\frac{hc}{\lambda_2} - \phi = \frac{2hc}{\lambda_1} - 2\phi

Bring the $\phi$ terms together:

-\phi + 2\phi = \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_2}

\phi = hc\left(\frac{2}{\lambda_1} - \frac{1}{\lambda_2}\right)

This is the clean algebraic result. Notice that the unknown $K_1$ cancelled out completely — that’s the power of using the ratio.

Plug in the numbers. Use $hc = 1240\ \text{eV·nm}$ (a standard and very convenient value for problems in eV and nm). …

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

←Previous questionQ13 — In the photoelectric effect, light (photons) is incident on a metal su…→Next questionQ15 — Assuming an electron is confined to a 1nm wide region, find the uncert…

Q.Consider a metal exposed to light of wavelength $600\ \text{nm}$ . The maximum energy of the electron doubles when light of wavelength $400\ \text{nm}$ is used. Find the work function in eV.

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: Solving Two-Wavelength Photoelectric Problems via Elimination

Steps

Q.Consider a metal exposed to light of wavelength 600 nm600\ \text{nm}600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm400\ \text{nm}400 nm is used. Find the work function in eV.

🎓Unlock everything free for 14 days

Q.Consider a metal exposed to light of wavelength $600\ \text{nm}$ . The maximum energy of the electron doubles when light of wavelength $400\ \text{nm}$ is used. Find the work function in eV.

Unlock everything free for 14 days