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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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If a stationary electron gets annihilated due to association with a stationary positron (hypothetically), the wavelength of resulting radiation will be .

$(a)\;\large\frac{h}{m_{0} C}\qquad(b)\;\large\frac{2h}{m_{0} C}\qquad(c)\;\large\frac{h}{2m_{0}C}\qquad(d)\;None$

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Answer : (a) $\;\large\frac{h}{m_{0} C}$
Explanation :
Where $m_{0}=\;$ rest mass of an electron (or positron)
C=speed of light
Considering the momenta of system just before and after event (anhiliation) we see that two identical photons will be resulted and travel in opposited directions with equal magnitude of momenta and energy $\;\large\frac{hc}{\lambda}\;$ where
$\lambda$= wavelength of radiation
Conservation of energy yields
$\large\frac{hC}{\lambda}+\large\frac{hC}{\lambda}=m_{0}C^2+m_{0}C^2$
$\large\frac{hC}{\lambda}=m_{0}C^2$
$\lambda=\large\frac{h}{m_{0}C}\;.$
answered Feb 25, 2014 by yamini.v
 

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