In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep=−(1+y)e where e is the electronic charge. Find the critical value of y such that expansion may start

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- let us consider that universe is of radius R

And we know that hydrogen is made up one electron and proton so net charge is

-(1+y)e+e = -ye

Now the number of hydrogen atom in the universe = N $\times \frac{4}{3} π R^{3}$

So total charge is = -ye $\times$ N $\times \frac{4}{3} π R^{3}$

...more

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- let us consider that universe is of radius R

And we know that hydrogen is made up one electron and proton so net charge is

-(1+y)e+e = -ye

Now the number of hydrogen atom in the universe = N $\times \frac{4}{3} π R^{3}$

So total charge is = -ye $\times$ N $\times \frac{4}{3} π R^{3}$

According to gauss theorem $? E . d s = \frac{q}{ε_{0}}$

So E $\times 4 π R^{2} = \frac{- y e \times N \times \frac{4}{3} π R^{3}}{ε_{0}}$

E= -NRye/3 $ε_{0}$

And we know electrostatic force F=qE = -ye(-NRye/3 ) $ε_{0}$ by using the value of E and q

F= $\frac{y^{2} e^{2} N R}{3 ε_{0}}$ force is positive which shows that force is repulsive.

But the gravity is g= $\frac{G M}{R^{2}}$ and gravitational force is F= $\frac{G M m}{R^{2}}$

But M= N $\times \frac{4}{3} π R^{3} m_{p}$ where $m_{p}$ is mass of proton

So using this mass in above eqn so force = $\frac{G N \times \frac{4}{3} π R^{3} m_{p} m}{R^{2}}$

But her masses of proton and hydrogen are equal so take as m

F= $\frac{G N \times 4 π R m^{2}}{3}$

So calculate its critical value electrostatic force and gravitational force both are equal.

So, $\frac{G N \times 4 π R m^{2}}{3} = \frac{y^{2} e^{2} N R}{3 ε_{0}}$

So y²= 4 $π ε_{0} \frac{m^{2}}{e^{2}}$ after using standard values we get value of y= approx. 10^-18

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This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation- let us consider that universe is of radius R             And we know that hydrogen is made up one electron and proton so net charge is             -(1+y)e+e = -ye             Now the number of hydrogen atom in the universe = N <math> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math>              So total charge is = -ye <math> <mo>×</mo> </math> N <math> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math>              According to gauss theorem <math> <mo>?</mo> <mi>E</mi> <mo>.</mo> <mi>d</mi> <mi>s</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math>               So E <math> <mo>×</mo> <mn>4</mn> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mrow> <mo>-</mo> <mi mathvariant="normal">y</mi> <mi mathvariant="normal">e</mi> <mo>×</mo> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">N</mi> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math>               E= -NRye/3 <math> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math>              And we know electrostatic force F=qE = -ye(-NRye/3 ) <math> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> by using the value of E and q             F= <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>N</mi> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> force is positive which shows that force is repulsive.             But the gravity is g= <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> and gravitational force is F= <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>              But M= N <math> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> </math>  where <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> </math> is mass of proton             So using this mass in above eqn so force = <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi mathvariant="normal">N</mi> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>              But her masses of proton and hydrogen are equal so take as m             F= <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi mathvariant="normal">N</mi> <mo>×</mo> <mn>4</mn> <mi>π</mi> <mi>R</mi> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> </math>              So calculate its critical value electrostatic force and gravitational force both are equal.             So, <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi mathvariant="normal">N</mi> <mo>×</mo> <mn>4</mn> <mi>π</mi> <mi>R</mi> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>N</mi> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math>               So y2= 4 <math> <mi>π</mi> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  after using standard values we get value of y= approx. 10-18

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