Question 492: Atoms & Nuclei

Question

For an electron moving in the $$\mathrm{n}^{	ext {th }}$$ Bohr orbit the deBroglie wavelength of an electron is

Accepted Answer

From de Broglie's hypothesis,
$$\lambda=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{n}}}=\frac{\mathrm{h}}{\mathrm{mv}} \quad 	ext{.... (i)}$$
and from Bohr's atomic model,
$$\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$$
Also,
$$\mathrm{L}=\mathrm{mvr}_{\mathrm{n}}$$
Due to quantization of angular momentum, we can write,
$$\begin{aligned}
& \frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}} \
& 	herefore \quad \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}} \quad 	ext{.... (ii)}
\end{aligned}$$
putting (ii) into (i), we get,
$$	herefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{m}(\mathrm{nh} / 2 \pi \mathrm{mr})}=\frac{2 \pi \mathrm{r}}{\mathrm{n}}$$

Answer

$$\mathrm{n} \pi \mathrm{r}$$

Answer

$$\frac{\pi r}{\mathrm{n}}$$

Answer

$$\frac{\mathrm{nr}}{2 \pi}$$

Answer

$$\frac{2 \pi r}{\mathrm{n}}$$

Atoms & Nuclei

For an electron moving in the $\mathrm{n}^{\text {th }}$ Bohr orbit the deBroglie wavelength of an electron is

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