Question

A schwarzschild black hole is characterized by its mass $M$ and a mathematical spherical surface of radius $R_{s} = \frac{2 G M}{C^{2}}$ called the event horizon. If the radial distance of an object $r$ from the black hole is such that $r < R_{s}$ , then the object is “swallowed” by the black hole and rapidly decreased to the singular point $r = 0$

(a) Suppose a black hole of mass $M$ “captures” a proton to form a “black hole proton atom $(B H P)$ ” in circular orbit. Find the smallest radius $r_{B}$ of this atom.
(b) Obtain a numerical upper bound on $M$ such that a stable $B H P$ may exist.
(c) Find the minimum energy $F_{\min}$ , in $MeV$ , required to dissociate this $B H P$ atom from the ground state.
(d) In $1974$ , Stephen Hawking showed that quantum effects cause black to radiate like a black body with temperature $T_{B H} = \frac{10^{23} K}{M}$ . Discuss the possibility of the existence of the stable $B H P$ atom.

Accepted Answer

Solution for (a):

Let's assume,

$m \to$ mass of a proton and

$v \to$ speed in the circular orbit.

As the black hole proton atom moves in a circular orbit about the black hole of mass $M$ .

$r < R_{s}$

$\Rightarrow v = \sqrt{\frac{G M}{r_{B}}} . . . (i)$

By Bohr's quantization principle,

$B H P$

$(∵ h = \frac{h}{2 π})$

$\Rightarrow v = \frac{n h}{2 π m r_{B}} . . . (ii)$

From equations $(i)$ and $(ii)$ ,

$\sqrt{\frac{G M}{r_{B}}} = \frac{n h}{m r_{B}}$

$\Rightarrow \frac{G M}{r_{B}} = {(\frac{n h}{m r_{B}})}^{2}$

$\Rightarrow r_{B} = \frac{n^{2} h^{2}}{m^{2} G M} . . . (ii)$

For smallest radius, consider $n = 1$ ,

$B H P$ .

Solution for (b):

For a stable $B H P$ to exist, $r_{B} > R_{s}$

$\Rightarrow \frac{h^{2}}{G M m^{2}} > \frac{2 G M}{c^{2}}$

$\Rightarrow M^{2} < \frac{c^{2} h^{2}}{2 G^{2} m^{2}}$

$\Rightarrow M < \frac{c h}{\sqrt{2} G m}$

$\Rightarrow M < \frac{(3 \times 10^{8}) (\frac{6.626 \times 10^{- 34}}{2 π})}{\sqrt{2} (6.67 \times 10^{- 11}) (1.67 \times 10^{- 27})}$

$\Rightarrow M < 2 \times 10^{11} kg$ .

Solution for (c):

Energy of the $B H P$ , $F = - \frac{G M m}{2 r}$

From $(ii)$ ,

$F = \frac{G^{2} M^{2} m^{3}}{2 n^{2} h^{2}}$

For $F_{\min}$ , $n = 1$ .

Thus, $F_{\min} = \frac{G^{2} M^{2} m^{3}}{2 h^{2}}$

$\Rightarrow F_{\min} = \frac{{(6.67 \times 10^{- 11})}^{2} {(2 \times 10^{11})}^{2} {(1.67 \times 10^{- 27})}^{3}}{2 {(\frac{6.626 \times 10^{- 34}}{2 π})}^{2} \times 1.6 \times 10^{- 19}} eV$

$\Rightarrow F_{\min} = 55 MeV$ .

Solution for (d):

$T_{B H} = \frac{10^{23} K}{M}$

For $M \approx 10^{11} kg$ ,

$\Rightarrow T_{B H} = 10^{12} K$ .

At this temperature, thermal energy is $82 MeV$ but the dissociation energy required is $55 MeV$ . Therefore the $B H P$ is thermally unstable.

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