If the force between the electron in the firs | vedclass

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Question

$\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{1}^{2}} ; \mathrm{F}_{2}=\frac{1}{4 \pi \varepsilon_{0}}

Vedclass · Accepted Answer

$F/16$

Vedclass · Answer

$\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{1}^{2}} ; \mathrm{F}_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{2}^{2}}$

Where $r_{2}$ is the radius of the second Bohr orbit

As $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^{2},$ so $\quad \mathrm{r}_{2}=4 \mathrm{r}_{1}$

$	herefore $ $\mathrm{F}_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left(4 \mathrm{r}_{1}ight)^{2}}=\frac{\mathrm{F}}{16}$

If the force between the electron in the first Bohr orbit and the nucleus (proton) in hydrogen atom is $F$, then the force between them when the electron is in the second orbit is

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