The Total solution for NCERT class 6-12
In an accelerator experiment on high-energycollisions of electrons with positrons, a certain event is interpreted asannihilation of an electron-positron pair of total energy 10.2 BeV into twoγ-rays of equal energy. What is the wavelength associated with each γ-ray?(1BeV = 109 eV)
Total energy of two γ-rays:
E = 10. 2 BeV
= 10.2 × 109 eV
= 10.2 × 109 × 1.6 × 10−10 J
Hence, the energy of each γ-ray:
Planck’s constant,
Speed of light,
Energy is related to wavelengthas:
Therefore, the wavelength associated witheach γ-ray is