In an accelerator experiment on high-energy collisions of electrons with positrons,
Question:

In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)

Solution:

Total energy of two γ-rays:

E = 10. 2 BeV

= 10.2 × 109 eV

= 10.2 × 109 × 1.6 × 10−10 J

$E^{\prime}=\frac{E}{2}$

$=\frac{10.2 \times 1.6 \times 10^{-10}}{2}=8.16 \times 10^{-10} \mathrm{~J}$

Planck’s constant, $h=6.626 \times 10^{-34} \mathrm{JS}$

Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Energy is related to wavelength as:

$E^{\prime}=\frac{h c}{\lambda}$

$\therefore \lambda=\frac{h c}{E^{\prime}}$

$=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{8.16 \times 10^{-10}}=2.436 \times 10^{-16} \mathrm{~m}$

Therefore, the wavelength associated with each $y$-ray is $2.436 \times 10^{-16} \mathrm{~m}$.

Hence, the energy of each γ-ray: