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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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

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प्रश्न

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.

  1. Find the critical value of y such that expansion may start.
  2. Show that the velocity of expansion is proportional to the distance from the centre.
दीर्घउत्तर
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उत्तर

a. Let the Universe have a radius of R. Assume that the hydrogen atoms are uniformly distributed. The charge on each hydrogen atom is

`e_H = – (1 + y) e + e = – ye = |ye|`

The mass of each hydrogen atom is ~ mp (mass of proton). Expansion starts if the Coulumb repulsion on a hydrogen atom, at R, is larger than the gravitational attraction. Let the Electric Field at R be E. Then

`4piR^2 E = 4/(3ε_0) pi R^3 N |ye|`  ....(Gauss's law)

E(R) = `1/3 (N|ye|)/ε_0 R hatr`

Let the gravitational field at R be GR. Then

`- 4piR^2 G_R = 4 piG  m_p  4/3 piR^3  N`

`G_R = - 4/3 pi Gm_pN R`

`G_R (R) = - 4/3 pi Gm_pN R hatr`

Thus the Coulombic force on a hydrogen atom at R is 

yeE(R) = `1/3 (Ny^2e^2)/ε_0 R hatr`

The gravitional force on this atom is

mpGr (R) = `- (4pi)/3 GNm_p^2 R hatr`

The net force on the atom is

F = `(1/3 (Ny^2e^2)/ε_0 R - (4pi)/3 GNm_p^2R)hatr`

The critical value is when

`1/3 (Ny^2e^2)/ε_0 R = (4pi)/3 GNm_p^2R`

⇒ `y_c^2 = 4piε_0 G m_p^2/e^2`

= `(7 xx 10^-11 xx 1.8^2 xx 10^6 xx 81 xx 10^-62)/(9 xx 10^9 xx 1.6^2 xx 10^-38)`

= 63 × 10–38

∴ yc = 8 × 10–19 = 10–18

b. Because of the net force, the hydrogen atom experiences an acceleration such that

`m_p (d^2R)/(dt^2) = (1/3 (Ny^2e^2)/e_o R - (4p)/3 GNm_p^2 R)`

Or, `(d^2R)/(dt^2) = a^2R` where `alpha^2 = 1/m_p (1/3 (Ny^2e^2)/e_o - (4p)/3 GNm_p^2)`

This has a solution R = `Ae^(at) + Be^(-at)`

As we are seeking an expansion, B = 0.

∴ R = Aeat

⇒ R = αAeat = αR

Thus, the velocity is proportional to the distance from the centre.

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पाठ 1: Electric Charges And Fields - MCQ I [पृष्ठ ७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
पाठ 1 Electric Charges And Fields
MCQ I | Q 1.26 | पृष्ठ ७

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