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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 $\gamma$ -rays of equal energy. What is the wavelength associated with each $\gamma$ -ray? $\left(1\; BeV =10^{9}\; eV \right)$
$6.254 \times 10^{-17}\; m$
$4.64 \times 10^{-15}\; m$
$2.436 \times 10^{-16}\; m$
$4.951 \times 10^{-15}\; m$
Solution
Total energy of two $\gamma$ -rays: $E =10.2 \,BeV$
$=10.2 \times 10^{9} eV$
$=10.2 \times 10^{9} \times 1.6 \times 10^{-10}\, J$
Hence, the energy of each $\gamma$ -ray
$E^{\prime}=\frac{E}{2}$
$=\frac{10.2 \times 1.6 \times 10^{-10}}{2}=8.16 \times 10^{-10}\, J$
Plank's constant, $h=6.626 \times 10^{-34}\, Js$
Speed of light $c=3 \times 10^{8} \,m / s$
Energy is related to wavelength as:
$E^{\prime}=\frac{h c}{\lambda}$
$\lambda=\frac{h c}{E^{\prime}}$
$=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{8.16 \times 10^{-10}}=2.436 \times 10^{-16} \,m$
Therefore, the wavelength associated with each $\gamma$ -ray is $2.436 \times 10^{-16}\; \,m$
