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Solution - Find the De Broglie Wavelength of a Neutron, in Thermal Equilibrium with Matter, Having an Average Kinetic Energy of (3/2) Kt at 300 K. - Wave Nature of Matter

ConceptWave Nature of Matter

Question

find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.

Solution

Temperature of the neutron, T = 300 K

Boltzmann constant, k = 1.38 × 10−23 kg m2 s−2 K−1

Average kinetic energy of the neutron:

K' = 3/2 kT

= 3/2 xx 1.38 xx 10^(-23) xx  300 = 6.21 xx 10^(-21) J

The relation for the de Broglie wavelength is given as:

lambda' = h/(sqrt(2K'm_n))

Where

m_n = 1.66 xx 10^(-27) Kg

h = 6.6 xx 10^(-34) Js

K' = 6.75 xx 10^(-21) J

:. lambda' = (6.63 xx 10^(-34))/sqrt(2xx 6.21 xx 10^(-21) xx 1.66 xx 10^(-27)) =  1.46 xx 10^(-10) m = 0.146 nm

Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

Is there an error in this question or solution?

APPEARS IN

NCERT Physics Textbook for Class 12 Part 2
Chapter 11: Dual Nature of Radiation and Matter
Q: 17.2 | Page no. 408

Reference Material

Solution for question: Find the De Broglie Wavelength of a Neutron, in Thermal Equilibrium with Matter, Having an Average Kinetic Energy of (3/2) Kt at 300 K. concept: Wave Nature of Matter. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
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