An electron is revolving around the nucleus of hydrogen atom. If $F$ is centripetal force on the electron in $2 n d$ orbit then the centripetal force on it in $1 s t$ orbit is:

F

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\frac{F}{4}

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16 F

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\frac{F}{16}

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Solution

The correct option is D $16 F$
$A c c o r d i n g t o e l e c t r o s t a t i c s f \propto \frac{1}{n^{4}} \Rightarrow \frac{f_{1}}{f_{2}} = \frac{n_{2}^{4}}{n_{1}^{4}} \Rightarrow f_{1} = 16 F$

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An electron is revolving around the nucleus of hydrogen atom. If F is centripetal force on the electron in 2 nd orbit then the centripetal force on it in 1 st orbit is:

The correct option is D 16FAccordingtoelectrostaticsf∝1n4⇒f1f2=n42n41⇒f1=16F

An electron is revolving around the nucleus of hydrogen atom. If $F$ is centripetal force on the electron in $2 n d$ orbit then the centripetal force on it in $1 s t$ orbit is:

The correct option is D $16 F$
$A c c o r d i n g t o e l e c t r o s t a t i c s f \propto \frac{1}{n^{4}} \Rightarrow \frac{f_{1}}{f_{2}} = \frac{n_{2}^{4}}{n_{1}^{4}} \Rightarrow f_{1} = 16 F$