91Ó°ÊÓ

We need to find the typical wavelength of the Hawking radiation emitted by a one-solar-mass.

Consider we have a black hole, the energy, temperature radius and surface area are given by respectively:

$u=m{c}^{2}T=\frac{h{c}^3}{16\mathrm{Ï€}kGM}r=\frac{2GM}{c^2}A=4\mathrm{Ï€}{r}^2$

The spectrum is given by:

$u\left(T\right)=\frac{8\mathrm{Ï€}}{{\left(hc\right)}^3}\frac{{\mathrm{Î\mu }}^3}{e^{\mathrm{Î\mu }/kT}-1}$

let $x=\mathrm{Î\mu }/kT$, then the spectrum will become

$u\left(T\right)=\frac{8\mathrm{Ï€}{\left(kT\right)}^3}{{\left(hc\right)}^3}\frac{x^3}{e^x-1}$

to find peak of the spectrum, we set the partial derivative with respect to energy (or x since the constant doesn't equal to zero) of the specturm equals to zero, that is

$\frac{a}{ax}\left(\frac{x^3}{e^x-1}\right))=0$

$\frac{3x^2(e^x-1)-x^3{e}^x}{{(e^x-1)}^2}=0$

$3x^2(e^x-1)-x^3{e}^x=0$

$3e^x-3-x{e}^x=0$

the solution of this equation is $x=2.8214$, which obtained using Matlab, and the code is shown in the following picture

$$\begin{gathered} >>fun=@\left(x\right)3*(exp(x)-1)-x,*exp\left(x\right); \\ >>x0=(13); \\ >>x=fzero(fun,x0) \\ x= \\ 2.8214 \end{gathered}$$

but,

$x=\frac{\mathrm{Î\mu }}{kT}=\frac{hc}{kT\mathrm{λ}}$where $\mathrm{Î\mu }=\frac{hc}{\mathrm{λ}}$

so the peak wavelength is:

$\mathrm{λ}=\frac{hc}{(2.8214)kT}$

substitute form (1) to get:

$\mathrm{λ}=\frac{hc}{(2.8214)k}\left(\frac{16{\mathrm{π}}^{2}kGM}{h{c}^3}\right)$

$\mathrm{λ}=\frac{16{\mathrm{π}}^{2}GM}{(2.8214)c^2}=\left(28\right)\frac{2GM}{c^2}$

$\mathrm{λ}=28r$

thus the peak wavelength equals 28 times the radius of the black hole or 14 times the diameter of the black hole

We need to calculate the total power radiated by a one-solar-mass black hole.

The irradiated power is given by:

$P=Ae\mathrm{σ}{T}^4$

where A is the area and e is the emissivity of the material. Substitute form (1) to get:

$P=4\mathrm{π}e\mathrm{σ}{\left(\frac{2GM}{c^2}\right)}^2{\left(\frac{h{c}^3}{16\mathrm{π}2kGM}\right)}^4$

$p=\frac{h{c}^6}{\left(30720\right){\mathrm{Ï€}}^6{G}^2{M}^2}$

substitute with the gives to get(note that l set e=1 since it is black body):

We need to obtain an expression for the lifetime of a black hole in terms of its initial mass.

The power equals the time derivative of the energy that is$\frac{dU}{dt}$, so we can write:

where the negative sign came from the fact that the mass of the black hole decreases with time. Substitute from get:

$\frac{d\left(M{c}^2\right)}{dt}=-\frac{h{c}^6}{\left(30720\right){\mathrm{Ï€}}^6{G}^2{M}^2}$

$\frac{dM}{dt}=-\frac{h{c}^4}{\left(30720\right){\mathrm{Ï€}}^6{G}^2{M}^2}$

$\frac{dM}{dt}=-\frac{C}{M^2}$

where,

$C=\frac{h{c}^4}{\left(30720\right){\mathrm{Ï€}}^6{G}^2}$

the value of this constant is:

$$\begin{gathered} C=\frac{(6.626×{10}^{-34}J×s){(3.00×{10}^{8}m/s)}^4}{\left(30720\right){\mathrm{π}}^6{(6.674×{10}^{-11}{m}^3/kg×s^2)}^2} \\ =4.08×{10}^{13}k{g}^3/s \end{gathered}$$

now we need to solve the differential equation as:

$M^{2}dM=-Cdt$

Integrate from the initial mass $M_i$to zero, and from o to the lifetime $\mathrm{Ï„}$to get:

$$\begin{gathered} {∫}_{M_1}^0{M}^{2}dM=-C{∫}_0^{T}dt \\ -\frac{{M_i}^3}{3}=-C_T \\ \mathrm{τ}=\frac{{M^3}_i}{3C} \end{gathered}$$

We need to calculate the lifetime of a one-solar-mass black hole.

For one solar mass black hole (the initial mass is $M_i=2.0×{10}^{30}kg)$the life time is:

$$\begin{gathered} \mathrm{τ}=\frac{{(2.0×{10}^{30}kg)}^3}{3(4.08×{10}^{13}k{g}^3/s} \\ =6.536×{10}^{76}s \\ =6.536×{10}^{76}s\left(\frac{1y}{365×24×60×{60}_s}\right) \\ =2.072×{10}^{69y} \\ \mathrm{τ}=2.072×{10}^{69}y \end{gathered}$$

which is more than ${10}^{59}$times the age of the known universe

We need to find the initial mass of the black hole and the electromagnetic spectrum would most of its radiation have been emitted.

The age of the known universe is $1.5×{10}^{10}y,or5×{10}^{17}s,$we need to find the mass of a black hole, such that it started with mass of $M_i$at the beginning of the universe and completely vaporized after the age of universe elapsed, so:

$M_i={\left(3C_{\mathrm{Ï„}}\right)}^{\frac{1}{3}}$

substitute with the givens to get:

$$\begin{gathered} M_i=\left(3\right(4.08×{10}^{13}k{g}^3/s){(5×{10}^{17}s))}^{\frac{1}{3}}=3.94×{10}^{10}kg \\ {M}_i=3.94×{10}^{10}kg \end{gathered}$$

to find what part of the spectrum that the most radiation have been emitted, we use the expression of part (a), so:

$\mathrm{λ}=\frac{56(6.674×{10}^{-11}{m}^3/kg×s^2)(3.94×{10}^{10}kg)}{{(3.0×{10}^{s}m/s)}^2}$

$$\begin{gathered} =1.636×{10}^{-15}m \\ \mathrm{λ}=1.636×{10}^{-15}m \end{gathered}$$