Q. 345.0( 3 Votes )

# The wavelength of

Answer :

Given:

Order of length of quark structure, λ = 10^{-15}m

Rest mass energy of electron = 0.511 MeV

i.e. m_{0}c^{2} = 0.511 MeV

→ m_{0}c^{2} = 0.511 × 1.6 × 10^{-19}

→ m_{0}c^{2} = 0.817 × 10^{-13} J

By using De-Broglie wavelength equation we can write,

Where,

p = momentum of particle

h = Planck’s constant = 6.6 × 10^{-34}Js

λ = wavelength

The equation of energy at relativistic speed is,

E^{2} = p^{2}c^{2} + m^{2}_{0}c^{4}

E^{2} = (6.6 × 10^{-19} × 3 × 10^{8})^{2} + (0.817 × 10^{-13})^{2}

→ E = (392.07 × 10^{-22})^{0.5}

→ E = 1.9 × 10^{-10}J

Hence the energy of electron emitted from linear accelerator is 1.9 × 10^{-10}J.

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