JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Modern Physics

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λ

and K are de Broglie Wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be :-

(A)Option (1)
(1)
(B)Option (2)
(2)
✓ Correct
(C)Option (3)
(3)
(D)Option (4)
(4)

Answer: (B)

Solution

The de Broglie wavelength of a particle of mass

m

and kinetic energy

K

λ = \frac{h}{p} = \frac{h}{2 m K}

λ^{2} = \frac{h ^{2}}{2 m} \cdot \frac{1}{K}

Plotting

1/ K

on the vertical axis against

λ

on the horizontal axis,

\frac{1}{K} = \frac{2 m}{h ^{2}} λ^{2}

This is a parabola opening upward and passing through the origin:

1/ K \to 0

λ \to 0

, and

1/ K

rises without bound as

λ

increases. This matches the curve that starts at the origin and curves upward.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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