MHT CET · Physics · Atomic Physics
For an electron moving in the \(\mathrm{n}^{\text {th }}\) Bohr orbit the deBroglie wavelength of an electron is
- A \(\mathrm{n} \pi \mathrm{r}\)
- B \(\frac{\pi \mathrm{r}}{\mathrm{n}}\)
- C \(\frac{\mathrm{n} \mathrm{r}}{2\pi}\)
- D \(\frac{2\pi \mathrm{r}}{\mathrm{n}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2\pi \mathrm{r}}{\mathrm{n}}\)
Step-by-step Solution
Detailed explanation
From de Broglie's hypothesis,
\(\lambda=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{n}}}=\frac{\mathrm{h}}{\mathrm{mv}}\)
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,
\(\begin{aligned}
& \frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}} \\
\therefore \quad & \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}
\end{aligned}\)
putting (ii) into (i), we get,
\(\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{m}(\mathrm{nh} / 2 \pi \mathrm{mr})}=\frac{2 \pi \mathrm{r}}{\mathrm{n}}\)
\(\lambda=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{n}}}=\frac{\mathrm{h}}{\mathrm{mv}}\)
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,
\(\begin{aligned}
& \frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}} \\
\therefore \quad & \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}
\end{aligned}\)
putting (ii) into (i), we get,
\(\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{m}(\mathrm{nh} / 2 \pi \mathrm{mr})}=\frac{2 \pi \mathrm{r}}{\mathrm{n}}\)
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