The characteristic length of entities in Superstring theory is approximately\({\rm{1}}{{\rm{0}}^{{\rm{ - 35}}}}{\rm{ m}}\). (a) Find the energy in GeV of a photon of this wavelength. (b) Compare this with the average particle energy of\({\rm{1}}{{\rm{0}}^{{\rm{19}}}}{\rm{ GeV}}\)needed for unification of forces.

The expression for the energy using the Planck’s quantum theory is given by,

\(E = h\frac{c}{\lambda }\)

Here\(E\)is the energy,\(h\)is the Planck’s constant,\(c\)is the speed of the light and\(\lambda \)is the wavelength.

(a)

The photon has a wavelength of \({\rm{1}}{{\rm{0}}^{{\rm{ - 35}}}}{\rm{ m}}\) would have energy of:

\(\begin{align}E &= \frac{{hc}}{\lambda }\\ &= \frac{{{\rm{6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{J \times 3 \times 1}}{{\rm{0}}^{\rm{8}}}\frac{{\rm{m}}}{{\rm{s}}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 35}}}}{\rm{m}}}}\\ &= {\rm{2 \times 1}}{{\rm{0}}^{{\rm{10}}}}{\rm{J}}\\ &= {\rm{1}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{20}}}}{\rm{GeV}}\end{align}\)

Therefore the energy of a photon is \({\rm{1}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{20}}}}{\rm{ GeV}}\).

(b)

Calculate the ratio of the energy of the photons to the average particles energy,

\(\begin{align}Ratio &= \frac{{1.2 \times {{10}^{20}}}}{{{{10}^{19}}}}\\ &= 12\end{align}\)

As here the energy of the photo is much greater than the average particles energy needed for unification of forces.