Question

If in Bohr's atomic model, it is assumed that force
between electron and proton varies inversely as r^4.
energy of the system will be proportional to
1.n^2
2.n^4
3.n^6
4.n^8​

Answer 1

Answer:

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Answer 2

The energy of the system will be proportional to n⁶.

(3) is correct option.

Explanation:

Given that,

The force between electron and proton varies inversely as r⁴.

We need to calculate the speed of electron

We know for electrons to move in the circular orbit  around the nucleus

{tex]K.E=P.E[/tex]

$\dfrac{mv^2}{r}=\dfrac{kZe^2}{r^4}$

Since force inversely proportional to $\dfrac{1}{r^4}$

$\dfrac{mv^2}{r}=\dfrac{k}{r^4}$

$v^2= \dfrac{k}{r^3}$

We need to calculate the radius

Using formula of Bohr model

$mvr=\dfrac{nh}{2\pi}$

Squaring both side

$m^2v^2r^2=\dfrac{n^2h^2}{4\pi^2}$

Put the value of v in to the formula

$m^2\times\dfrac{k}{r^3}\times r^2=\dfrac{n^2h^2}{4\pi^2}$

$r=\dfrac{m^24\pi^2 k}{n^2 h^2}$

We need to calculate the energy

Using formula of kinetic energy

$E =K.E$

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}m\times\dfrac{k}{r^3}$

Put the value of r into the formula

$E=\dfrac{1}{2}m\times\dfrac{k}{(\dfrac{m^2\times4\pi^2\times k}{n^2 h^2)^3}}$

$E=\dfrac{1}{2}\times\dfrac{n^6\times h^6}{m\times4\pi^6}$

$E\propto n^6$

Hence, The energy of the system will be proportional to n⁶.

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Topic : Bohr model

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