If the radius of first Bohr orbit is $^{'} r^{'}$ , then radius of second orbit will be

2 r

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\frac{r}{2}

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4 r

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\sqrt{2 r}

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Solution

The correct option is B $4 r$
$r = \frac{E_{0} n^{2} h^{2}}{R m z e^{2}}$
So, $r \propto n^{2}$
So, $\frac{r_{1}}{r_{2}} = (\frac{n_{1}}{n_{2}})^{2}$
$\Rightarrow \frac{r}{r_{2}} = (\frac{n}{2 n})^{2}$ ( $∵$ second orbit has n=2)
$\Rightarrow r_{2} = 4 r$

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If the radius of first Bohr orbit is ′r′, then radius of second orbit will be

The correct option is B 4rr=E0n2h2Rmze2So, r∝n2So, r1r2=(n1n2)2⇒rr2=(n2n)2 (∵second orbit has n=2)⇒r2=4r

If the radius of first Bohr orbit is $^{'} r^{'}$ , then radius of second orbit will be

The correct option is B $4 r$
$r = \frac{E_{0} n^{2} h^{2}}{R m z e^{2}}$
So, $r \propto n^{2}$
So, $\frac{r_{1}}{r_{2}} = (\frac{n_{1}}{n_{2}})^{2}$
$\Rightarrow \frac{r}{r_{2}} = (\frac{n}{2 n})^{2}$ ( $∵$ second orbit has n=2)
$\Rightarrow r_{2} = 4 r$