Question

How much energy must a gamma ray photon have, if it is to materialize into a pair of electron and positron, with each particle having a $K.E$ of $1\text{MeV}$?

• A. $2\text{MeV}$
• B. $1.0\text{MeV}$
• C. $0.5\text{MeV}$
• D. $3.02\text{MeV}$

Answer 1

The correct option is D $3.02\text{MeV}$

By conservation of energy,

$E_{gamma}=E_{\text{electron}}+E_{\text{positron}}+2K.E$

Mass of electron and positron are same,

$M_e=M_p=9.1\times {10}^{-31}\text{kg}$

By, Einstein's Mass-Energy relation,

$E=M{c}^2$

$\Rightarrow E_{\text{electron}}=E_{\text{positron}}=9.1\times {10}^{-31}\times (3\times {10}^8{)}^2$

$\Rightarrow E_{\text{electron}}=9.1\times 9\times {10}^{-15}\text{J}$

$\Rightarrow E_{\text{electron}}=\frac{81.9\times {10}^{-15}}{1.6\times {10}^{-13}}\text{MeV}$

$\Rightarrow E_{\text{electron}}=0.512\text{MeV}$

$\Rightarrow E_{gamma}=(2\times 0.51\text{MeV})+(2\times 1\text{MeV})$

$\Rightarrow E_{gamma}=3.02\text{MeV}$

Hence, option $\left(D\right)$ is correct.