11.24 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 -rays of equal energy. What is the wavelength associated with each -ray? (1BeV = 109 eV)
11.24 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 -rays of equal energy. What is the wavelength associated with each -ray? (1BeV = 109 eV)
11.24 The total energy of two X-rays = 10.2 BeV = 10.2 = 10.2 J
Hence energy of each X-ray = = 8.16 J
Planck's constant, h = 6.626 Js
Speed of light, c = 3 m/s
From the relation of energy and wavelength, we get
= or
= = 2.436 m
Therefore the wavelength associated with each X-ray is 2.436 m
Similar Questions for you
Based on theory
z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV
- (i)
- (ii)
from (i) & (ii)
ev
hu = hu0 + K.E
Cases u = 2u0
h2u0 = hu0 + K.E1
K.E1 = hu0
- (1)
Now, cases 2
h 5u0 = hu0 + k.E2
k.E2 = 4hu0
v2 =
v2 = 2v1
This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:
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Physics Ncert Solutions Class 12th 2023
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