Physics · Dual Nature Of Radiation And Matter · NEET
The de Broglie wavelength (lambda) vs momentum (p) graph is a rectangular hyperbola. This is because lambda = h/p, so lambda is inversely proportional to p: as p goes up, lambda goes down, and the curve never touches either axis. Memory hook: "h stays on top, so lambda and p make a see-saw" - one rises, the other falls, tracing a smooth downward curve.
Left: lambda vs p is a rectangular hyperbola because lambda = h/p (inverse). Right: 1/lambda vs p is a straight line through the origin with slope 1/h (direct). Knowing which is which is the key to the NEET graph questions.
Your doubts, answered
Is the de Broglie wavelength vs momentum graph a straight line or a curve?
It is a curve, not a straight line. The relation is lambda = h/p. Here h (Planck's constant) is fixed, so lambda = (constant)/p. This is an inverse relation, which always plots as a rectangular hyperbola. A straight line would mean lambda is directly proportional to p, which is wrong.
Why is it a hyperbola and not a straight line through the origin?
Because lambda times p = h = constant. Whenever the product of two quantities is a fixed number, their graph is a rectangular hyperbola. If it were a straight line through the origin, that would mean lambda = (constant) times p (direct proportion). But doubling the momentum halves the wavelength, so it is the opposite: inverse proportion.
Does the lambda vs p curve ever touch the axes?
No. As p becomes very large, lambda gets very small but never reaches zero (it just approaches the p-axis). As p becomes very small, lambda gets very large but never reaches infinity on the graph (it approaches the lambda-axis). Both axes are asymptotes - lines the curve gets close to but never meets.
What does the 1/lambda vs p graph look like?
A straight line through the origin. Since lambda = h/p, we get 1/lambda = p/h = (1/h) times p. So 1/lambda is directly proportional to p, giving a straight line passing through the origin with slope 1/h. This is a common way NET reshapes the same idea.
What is the (1/lambda)^2 vs kinetic energy E graph? (NEET 2024)
A straight line through the origin. For a free particle, p = sqrt(2mE) and lambda = h/p, so (1/lambda)^2 = p^2/h^2 = (2m/h^2) times E. Thus (1/lambda)^2 is directly proportional to E - a straight line from the origin with slope 2m/h^2. The heavier the particle, the steeper the line.
⚠️ The NEET trap ✗ Picking a straight line for the lambda vs p graph, thinking 'wavelength grows with momentum'. ✓ lambda = h/p means lambda is INVERSELY proportional to p, so the correct graph is a rectangular hyperbola that never touches the axes. 🧠 Straight line = direct proportion. Hyperbola = inverse proportion. lambda and p are inverse, so it must curve down.
Real NEET questions
2022
The graph which shows the variation of the de Broglie wavelength (lambda) of a particle and its associated momentum (p) is:
A · A rising straight line through the origin
B · A horizontal straight line
C · A parabola opening upward
D · A rectangular hyperbola (lambda decreases as p increases) ✓
Solution: de Broglie relation: lambda = h/p, so lambda is proportional to 1/p. When two quantities have a constant product (lambda times p = h), the graph is a rectangular hyperbola. lambda falls as p rises and the curve never touches either axis. So the correct option is the hyperbola.
2024
The graph which shows the variation of (1/lambda)^2 and the kinetic energy E of a free particle (where lambda is the de Broglie wavelength) is:
A · A hyperbola
B · A straight line with a negative slope
C · A straight line through the origin ✓
D · A horizontal line
Solution: Step 1: lambda = h/p and for a free particle E = p^2/2m, so p = sqrt(2mE). Step 2: 1/lambda = p/h, so (1/lambda)^2 = p^2/h^2. Step 3: substitute p^2 = 2mE to get (1/lambda)^2 = (2m/h^2) times E. Since 2m/h^2 is constant, (1/lambda)^2 is directly proportional to E - a straight line passing through the origin.
Solved Dual Nature Of Radiation And Matter NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the equation behind the de Broglie wavelength vs momentum graph?
lambda = h/p, where h = 6.63 x 10^-34 J s is Planck's constant and p is momentum. This can be written as lambda times p = h (a constant), which is the equation of a rectangular hyperbola.
What is the shape of the lambda vs p graph?
A rectangular hyperbola in the first quadrant. lambda decreases as p increases, and the curve approaches both axes but never touches them.
How is lambda related to momentum - directly or inversely?
Inversely. lambda is proportional to 1/p. Doubling the momentum halves the de Broglie wavelength.
Which graph is a straight line - lambda vs p or 1/lambda vs p?
1/lambda vs p is a straight line through the origin with slope 1/h. lambda vs p is a hyperbola. NEET often swaps these to test whether you know the difference between direct and inverse proportion.
Why does the (1/lambda)^2 vs E graph pass through the origin?
Because (1/lambda)^2 = (2m/h^2) times E, a direct proportion with no constant term. When E = 0, (1/lambda)^2 = 0, so the line starts exactly at the origin.
