Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.31
NCERT Solutions for Class 12 Physics Chapter 11 Question-31
Additional Exercise
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me=9.11 × 10⁻³¹ kg).
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An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Aº = 10⁻¹⁰ m
Mass of an electron, me = 9.11 x 10-31 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Charge on an electron, e = 1.6 x 10-19 C
The kinetic energy of the electron is given as:
E = 1/2 mev²
=>mev = √ (2Eme)
Where,
v = Velocity of the electron
mev = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
λ = h/p = h/mev = h/√ (2Eme)
Therefore , E = h²/(2λ²me) 2
= (6.6 x 10⁻34)² /2 x ( 10⁻¹⁰)² x(9.11 x 10-31 ) = 2.39 x 10⁻¹⁷ J
= ( 2.39 x 10⁻¹⁷ )/(1.6 x 10-19) = 149 .375 eV
Energy of a photon,E’ =hc/λ e eV
= (6.6 x 10⁻34) x (3 x 10⁸ ) /( 10⁻¹⁰) (1.6 x 10-19)
= 12.375 x 10³ eV =12.375 keV
Hence, a photon has a greater energy than an electron for the same wavelength.