In a hypothetical model of an atom following Bohr's theory, the potential energy is given as P.E. = -−Ke24r4. Which of the following option is correct? (Symbols have usual meaning)

We know that;PE=−Ke24r4 Force =−dEdr=−4Ke24r5=mv2rmvr=nh2πke2r5=mrn2h24π2m2r2r2=4π2mke2n2h2 ⇒r=2πenhmkThus,v=nh2πm⋅nh2πemk=n2h24π2m3/2ek

In a hypothetical model of an atom following Bohr's theory, the potential energy is given as P.E. = -−Ke24r4. Which of the following option is correct? (Symbols have usual meaning)

We know that;PE=−Ke24r4 Force =−dEdr=−4Ke24r5=mv2rmvr=nh2πke2r5=mrn2h24π2m2r2r2=4π2mke2n2h2 ⇒r=2πenhmkThus,v=nh2πm⋅nh2πemk=n2h24π2m3/2ek

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In a hypothetical model of an atom following Bohr's theory, the potential energy is given as P.E. = - $- \frac{{Ke}^{2}}{4 r^{4}}$ . Which of the following option is correct?
(Symbols have usual meaning)

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$r = \frac{nhe}{4 π} \sqrt{\frac{K}{m}}$

$v = \frac{n^{2} h^{2}}{4 π^{2} m^{3 / 2} e \sqrt{K}}$

$TE = \frac{- {Ke}^{2}}{2 r^{4}}$

$KE = - PE$

answer is C.

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Detailed Solution

We know that;

$PE = - \frac{{Ke}^{2}}{4 r^{4}}$

$Force = - \frac{dE}{dr} = - \frac{4 {Ke}^{2}}{4 r^{5}} = \frac{{mv}^{2}}{r}$

$mvr = \frac{nh}{2 π}$

$\frac{{ke}^{2}}{r^{5}} = \frac{m}{r} (\frac{n^{2} h^{2}}{4 π^{2} m^{2} r^{2}})$

$r^{2} = \frac{4 π^{2} {mke}^{2}}{n^{2} h^{2}} \Rightarrow r = \frac{2 πe}{nh} \sqrt{mk}$

Thus,

$v = \frac{nh}{2 πm} \cdot \frac{nh}{2 πe \sqrt{mk}} = \frac{n^{2} h^{2}}{4 π^{2} m^{3 / 2} e \sqrt{k}}$

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In a hypothetical model of an atom following Bohr's theory, the potential energy is given as P.E. = -−Ke24r4. Which of the following option is correct? (Symbols have usual meaning)

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In a hypothetical model of an atom following Bohr's theory, the potential energy is given as P.E. = - $- \frac{{Ke}^{2}}{4 r^{4}}$ . Which of the following option is correct?
(Symbols have usual meaning)