A photon of 1.7 times 10^{-13} J is absorbed | vedclass

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Question

(B) For electron-positron pair production,the minimum energy required is $1.02 \ MeV$.The energy of the given photon is $E = 1.7 	imes 10^{-13} \ J$.

Vedclass · Accepted Answer

Electron and positron pair will be created.

Vedclass · Answer

(B) For electron-positron pair production,the minimum energy required is $1.02 \ MeV$.The energy of the given photon is $E = 1.7 	imes 10^{-13} \ J$.To convert this energy into $MeV$,we divide by the charge of an electron $(1.6 	imes 10^{-19} \ C)$:$E = \frac{1.7 	imes 10^{-13}}{1.6 	imes 10^{-19}} \ eV = 1.0625 	imes 10^6 \ eV = 1.0625 \ MeV$.Since the energy of the photon $(1.0625 \ MeV)$ is greater than the threshold energy required for pair production $(1.02 \ MeV)$,an electron-positron pair will be created.

$A$ photon of $1.7 \times 10^{-13} \ J$ is absorbed by a material under special circumstances. The correct statement is:

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