The P.E. 'U' of a moving particle of mass 'm' varies with 'x'-axis as shown in figure. The deBroglie wavelength or the particle in the regions 0 ≤ x ≤ 1 and x > 1 are λ₁ and λ₂ respectively. II the total energy of the particle is ' nE ', then the ratio λ₁/λ₂ is (a) √((n²)/(n-1)) (b) √(n-1/n) (c) √(n/n-1) (d) √((n(n-1))/(n))

Answer: (c). Total energy nE; kinetic energy K = nE - U. 0 ≤ x ≤ 1: U = E ⇒ K₁ = (n-1)E; x > 1: U = 0 ⇒ K₂ = nE λ = (h)/(√(2mK)) ⇒ (λ₁)/(λ₂) = √((K₂)/(K₁)) = √(n/n-1) So the answer is (c) √(n/n-1).

The P.E. 'U' of a moving particle of mass 'm' varies with 'x'-axis as… | MHT CET 2021 (20 Sep, Shift 2)

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Quantum Numbers, X-rays & Other Questions

The P.E. 'U' of a moving particle of mass 'm' varies with 'x'-axis as shown in figure. The deBroglie wavelength or the particle in the regions $0 \le x \le 1$ and $x > 1$ are $\lambda_1$ and $\lambda_2$ respectively. II the total energy of the particle is ' $nE$ ', then the ratio $\lambda_1/\lambda_2$ is

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Answer: (c)

Total energy $nE$ ; kinetic energy $K = nE - U$ .
$0 \le x \le 1$ : $U = E \Rightarrow K_1 = (n-1)E$ ; $x > 1$ : $U = 0 \Rightarrow K_2 = nE$
$\lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{n}{n-1}}$
So the answer is (c) $\sqrt{\frac{n}{n-1}}$ .

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