Question 512: Atoms & Nuclei

Question

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 $$0 \leq x \leq 1$$ and $$x > 1$$ are $$\lambda_1$$ and $$\lambda_2$$ respectively. II the total energy of the particle is '$$\mathrm{nE}$$', then the ratio $$\lambda_1 / \lambda_2$$ is

Accepted Answer

In the region $$0 \leq x \leq 1$$, the potential energy of the particle is $$E$$.
Total energy is $$\mathrm{nE}$$.
Hence, kinetic energy, $$K=n E-E=(n-1) E$$
Its momentum, $$\mathrm{p}_1=\sqrt{2 \mathrm{mK}}=\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}$$
$$	herefore \lambda_1=\frac{\mathrm{h}}{\mathrm{p}_1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}}$$
In the region $$x > 1, \mathrm{PE}$$ is zero.
Hence, total energy is kinetic energy.
$$\begin{aligned}
& 	herefore \mathrm{K}=\mathrm{nE} \
& 	herefore \lambda_2=\frac{\mathrm{h}}{\mathrm{p}_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \
& 	herefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}}
\end{aligned}$$

Answer

$$\sqrt{\frac{n^2}{n-1}}$$

Answer

$$\sqrt{\frac{n-1}{n}}$$

Answer

$$\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}}$$

Answer

$$\sqrt{\frac{\mathrm{n}(\mathrm{n}-1)}{\mathrm{n}}}$$

Atoms & Nuclei

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