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MHT-CET Physics · Formula sheet

Dual Nature of Radiation and Matter formulas

7 formulas and 12 common traps for MHT-CET Physics Dual Nature of Radiation and Matter, grouped by subtopic.

Full notes with worked examples

The Photoelectric Effect

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What Intensity and Frequency Each Control

Threshold

hν0=ϕ=hcλ0h\nu_0 = \phi = \frac{hc}{\lambda_0}

Einstein's Equation in One Situation

Einstein's equation

hν=ϕ+12mvmax⁡2=ϕ+eV0h\nu = \phi + \tfrac{1}{2}mv_{\max}^2 = \phi + eV_0

Threshold From Two Readings

Two readings

e(V1−V2)=hc(1λ1−1λ2),λ0=hcϕe(V_1 - V_2) = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right), \qquad \lambda_0 = \frac{hc}{\phi}

Reading the Photoelectric Graphs

Stopping-potential line

V0=heν−ϕeV_0 = \frac{h}{e}\nu - \frac{\phi}{e}

Common traps

Letting intensity change the stopping potential

Brighter light means more photons of the SAME energy. The number of electrons goes up; their maximum energy, and so the stopping potential, does not.

Saying doubled frequency doubles the kinetic energy

KE = hν − φ. At 2ν it is 2hν − φ = 2KE + φ — more than double. The stopping potential likewise becomes more than double.

Taking the speed ratio from the photon energies

Speed goes with the kinetic energy, and KE is the photon energy MINUS φ. Photons of 2φ and 3φ give KE φ and 2φ, speeds 1 : √2 — not √2 : √3.

Scaling the work function with the frequency

φ belongs to the metal. Two metals with φ in the ratio 1 : 2 lit by f and 2f give KE hf − φ and 2hf − 2φ, a ratio of 1 : 2.

Dividing the two equations directly

V₁/V₂ is not (hc/λ₁)/(hc/λ₂): φ sits in both. Multiply one equation so the stopping potentials match, then subtract.

Swapping λ₁ and λ₂ in the answer

The longer wavelength gives the SLOWER electron. With speeds V at λ₁ and 2V at λ₂, the 4 goes with 1/λ₁: φ = (hc/3)(4/λ₁ − 1/λ₂).

Reading intensity from the cut-off

The cut-off on the potential axis is set by frequency; the saturation height is set by intensity. Curves that meet at one stopping potential share a frequency, whatever their heights.

The de Broglie Wavelength

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Wavelength, Momentum and Kinetic Energy

de Broglie wavelength

λ=hp=h2mE\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}

Electrons Accelerated Through a Potential Difference

Accelerated electron

λ=h2meV≈12.27V A˚\lambda = \frac{h}{\sqrt{2meV}} \approx \frac{12.27}{\sqrt{V}}\ \text{Å}

The Electron's Wave Round a Bohr Orbit

Standing wave on an orbit

nλ=2πrn,λn=2πnr1n\lambda = 2\pi r_n, \qquad \lambda_n = 2\pi n r_1

Common traps

Writing λ ∝ 1/E

Momentum goes as the square root of kinetic energy, so λ ∝ E^(−1/2). Doubling E divides λ by √2, not by 2.

Using λ = h/√(2mE) for a photon

A photon has no rest mass; its wavelength is hc/E. Comparing a photon and a particle means using a different formula for each.

Reading 'decreased to 1/√2 times' as 'increased'

A larger voltage gives a larger momentum and a SHORTER wavelength. The options pair the right factor with the wrong direction.

Picking the steepest line as the heaviest particle

On λ against 1/√V, the slope is h/√(2mq): a heavier particle has a SMALLER slope.

Dividing the first-orbit circumference by n

λ = 2πrₙ/n uses the radius of the nth orbit, n²r. With r₁ it gives 2πr/n, which shrinks with n; the correct λ = 2πnr grows.

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