Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.24
NCERT Solutions for Class 12 Physics Chapter 11 Question-24
Additional Exercise
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 10⁹ eV)
Share
Total energy of two y-rays: E = 10. 2 BeV = 10.2 x 109 eV = 10.2 x 109 x 1.6 x 10⁻¹⁹ J
Hence, the energy of each γ-ray:
E’ = E/2 = (10.2 x 10⁻¹⁰)/2 = 8.16 x 10⁻¹⁰ J
Planck’s constant, h = 6.626 x 10 ⁻³4 Js
Speed of light, c = 3 x 10⁸ m/s
Energy is related to wavelength as:
E’ = hc/λ
Therefore , λ = hc/E’
= (6.626 x 10⁻34 x 3 x 10⁸)/( 8.16 x 10⁻¹⁰)
= 2.436 x 10-16 m
Therefore, the wavelength associated with each γ-ray is 2.436 x 10-16 m.