Question

When a positron and an electron collide, they are annihilated and two gamma photons of equal energy are emitted. Calculate the wavelength corresponding to this gamma emission:
Plan Energy per photon$=m_e{C}^2$
$E=hv=\frac{hc}{\lambda }$
Thus, $\lambda$ can be calculated.

• A. $2.42$ pm
• B. $4.52$ pm
• C. $3.42$ pm
• D. None of these

Answer 1

The correct option is A $2.42$ pm

${}_1{e}^0{+}_{-1}{e}^0⟶2\gamma$ photos of equal energy
The energy of the above process is
$E=2m_e{C}^2$
$=2(9.11\times {10}^{-31}kg)(3\times {10}^{8}m{s}^{-1}{)}^2$
$=1.6398\times 10-13$ J
$\therefore$ energy of one photon$=8.199\times {10}^{-14}$ J
taking $E=\frac{hc}{\lambda }$
$\lambda =\frac{hc}{E}=\frac{6.62\times {10}^{-34}\times 3\times {10}^{8}m}{8.199\times 10-14}$ m
$=2.42\times {10}^{-12}$ $m=2.42$ pm