Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.17
NCERT Solutions for Class 12 Physics Chapter 11 Question-17
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10⁻¹⁰ m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K
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Ans (a).
De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m
Mass of a neutron, mn = 1.66 x 10-27 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Kinetic energy (K) and velocity (v) are related as :
K = 1/2 mnv2 —————–Eq-1
De Broglie wavelength (λ) and velocity (v) are related as :
λ = h/mnv—————–Eq-2
Using Eq-2 in Eq-1 .we get :
K = 1/2 x (mnh²)/(λ²m²n) = h²/2 λ²mn
=(6.6 x 10⁻34)² /2 (1.40 x 10⁻10)² (1.66 x 10-27)
= 6.75 x 10⁻²¹ J
= (6.75 x 10⁻²¹) /(1.6 x 10⁻¹⁹ )
= 4.219 x 10⁻² eV
Hence ,the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV
Ans (b).
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10-23 kg m2 s⁻2 K⁻1
Average kinetic energy of the neutron:
K’ = 3/2 k T = 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J
The relation for the de Broglie wavelength is given as:
λ’ =h/ √ (2K’mn)
Where, mn = 1.66 x 10-27 kg, h = 6.6 x 10-34 Js and K‘ = 6.75 x 10-21 J.
Therefore,
λ’ = (6.6 x 10-34)/√[2 x (6.21 x 10-21) x (1.66 x 10-27) ]
= 1.46 x 10–10 m = 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.