Question:medium
The graph which shows the variation of \(\big(\frac{1}{\lambda^2} \big)\)and its kinetic energy, is \(E\) (where \(λ\) is de Broglie wavelength of a free particle):
The graph which shows the variation of \(\big(\frac{1}{\lambda^2} \big)\)and its kinetic energy, is \(E\) (where \(λ\) is de Broglie wavelength of a free particle):
Updated On: Jan 13, 2026
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The Correct Option is D
Solution and Explanation
Step 1: De Broglie Wavelength Formula Recall
The de Broglie wavelength is defined as:
λ = h / √(2mE)
- λ represents the de Broglie wavelength.
- E denotes kinetic energy.
- m signifies mass.
- h is Planck’s constant.
Step 2: Deriving 1/λ² in Relation to E
Squaring the formula yields:
λ² = h² / 2mE
Inverting this expression gives:
1/λ² = 2mE / h²
This demonstrates that 1/λ² is directly proportional to E.
Step 3: Graph Interpretation
The relationship between 1/λ² and E is graphically represented by a straight line that passes through the origin.
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