MHT CET · Physics · Dual Nature of Matter
If the radius of the first Bohr orbit is ' \(r\) ' then the de-Broglie wavelength of the electron in the \(4^{\text {th }}\) orbit will be
- A \(4 \pi r\)
- B \(6 \pi r\)
- C \(8 \pi r\)
- D \(\frac{\pi \mathrm{r}}{4}\)
Answer & Solution
Correct Answer
(C) \(8 \pi r\)
Step-by-step Solution
Detailed explanation
According to Bohr's second postulate, \(\frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}}\)
\(\therefore\) de-Broglie wavelength, \(\lambda_{\mathrm{n}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}_{\mathrm{n}}}{\mathrm{n}}\)
Also, \(r_n \propto n^2\)
\(\therefore\) The de-Broglie wavelength of the electron in the \(4^{\text {th }}\) orbit is:
\(
\begin{aligned} \lambda_4 & =\frac{2 \pi r_4}{4}=\frac{2 \pi \times(16 r)}{4} \\
\therefore \lambda_4 & =8 \pi \mathrm{r}
\end{aligned}
\)
\(\therefore\) de-Broglie wavelength, \(\lambda_{\mathrm{n}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}_{\mathrm{n}}}{\mathrm{n}}\)
Also, \(r_n \propto n^2\)
\(\therefore\) The de-Broglie wavelength of the electron in the \(4^{\text {th }}\) orbit is:
\(
\begin{aligned} \lambda_4 & =\frac{2 \pi r_4}{4}=\frac{2 \pi \times(16 r)}{4} \\
\therefore \lambda_4 & =8 \pi \mathrm{r}
\end{aligned}
\)
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