Q.Monochromatic radiation of wavelength $640.2\ \text{nm}$ ($1\text{nm} = 10^{-9}\ \text{m}$) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be $0.54\ \text{V}$. The source is replaced by an iron source and its $427.2\ \text{nm}$ line irradiates the same photo-cell. Predict the new stopping voltage.

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:

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

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:

2. In Simple Harmonic Motion (SHM)

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

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:

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

$v_{\text{max}} = A\omega$ …