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प्रश्न
The graph which shows the variation of `(1/lambda^2)` and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
पर्याय
MCQ
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उत्तर
Explanation:
We have λ = `h/(sqrt(2 m E))`
⇒ `sqrt E prop 1/lambda`
⇒ E ∝ `1/lambda^2`
∴ The graph passes through the origin with a constant slope.
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