MCQ
If ratio of Area of two orbits of $H$ atom is $4 : 1$ then the ratio of frequency of $e^-$ in these two orbits is
- A$\frac{8}{1}$
- B$\frac{{2\sqrt 2 }}{1}$
- ✓$\frac{1}{{2\sqrt 2 }}$
- D$\frac{1}{8}$
✓
Answer
Correct option: C.
$\frac{1}{{2\sqrt 2 }}$
c
$\frac{A_{1}}{A_{2}}=\frac{4}{1}=\left(\frac{n_{1}}{n_{2}}\right)^{4}$
$\frac{A_{1}}{A_{2}}=\frac{4}{1}=\left(\frac{n_{1}}{n_{2}}\right)^{4}$
$\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\left(\frac{4}{1}\right)^{\frac{1}{4}}=\sqrt{2}$
Freq. $\propto \frac{1}{n^{3}}$
$\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)^{3} \quad \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{1}{\sqrt{2}}\right)^{3}$
$\frac{r_{1}}{r_{2}}=\frac{1}{2 \sqrt{2}}$
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