The ratio of centripetal forces on two electrons which are revolving around the nucleus of hydrogen atom in $2 n d,$ 3\ rd$ orbits respectively is:

27 : 8

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81 : 16

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8 : 27

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16 : 81

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Solution

The correct option is B $81 : 16$
Radius of Bohr's orbit $\propto n^{2}$
Velocity in Bohr's orbit $\propto \frac{1}{n}$
$\Rightarrow$ Centripetal force $= \frac{m ν^{2}}{r} \propto \frac{1}{n^{4}}$
$\Rightarrow F_{1} : F_{2} = (3 : 2)^{4} = 81 : 16$

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The ratio of centripetal forces on two electrons which are revolving around the nucleus of hydrogen atom in 2 nd,3\ rd$ orbits respectively is:

The correct option is B 81:16Radius of Bohr's orbit ∝n2Velocity in Bohr's orbit ∝1n⇒ Centripetal force =mν2r∝1n4⇒F1:F2=(3:2)4=81:16

The ratio of centripetal forces on two electrons which are revolving around the nucleus of hydrogen atom in $2 n d,$ 3\ rd$ orbits respectively is:

The correct option is B $81 : 16$
Radius of Bohr's orbit $\propto n^{2}$
Velocity in Bohr's orbit $\propto \frac{1}{n}$
$\Rightarrow$ Centripetal force $= \frac{m ν^{2}}{r} \propto \frac{1}{n^{4}}$
$\Rightarrow F_{1} : F_{2} = (3 : 2)^{4} = 81 : 16$