The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

  • Option 1)

    \frac{\text{h}} {{\sqrt {\text{mkT}} }}

  • Option 2)

    \frac{\text{h}} {{\sqrt {\text{3mkT}} }}

  • Option 3)

    \frac{{\text{2h}}} {{\sqrt {\text{3mkT}} }}

  • Option 4)

    \frac{{\text{2h}}} {{\sqrt {\text{mkT}} }}

 

Answers (1)
P Prateek Shrivastava

As we learnt in 

De - Broglie wavelength with charged particle -

lambda = frac{h}{sqrt{2mE}}= frac{h}{sqrt{2mE}}

lambda = frac{h}{sqrt{2mqv}}
 

- wherein

E
ightarrow kinetic: energy: o! f particle

q
ightarrow charged : particle

 

 Kinetic energy of neutron is 

K.E=\frac{3}{2}KT(Due to thermal motion)

=> \frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2m.\left ( \frac{3}{2}KT \right )}}

\Rightarrow \lambda =\frac{h}{\sqrt{3mkT}}


Option 1)

\frac{\text{h}} {{\sqrt {\text{mkT}} }}

Option is incorrect

Option 2)

\frac{\text{h}} {{\sqrt {\text{3mkT}} }}

Option is correct

Option 3)

\frac{{\text{2h}}} {{\sqrt {\text{3mkT}} }}

Option is incorrect

Option 4)

\frac{{\text{2h}}} {{\sqrt {\text{mkT}} }}

Option is incorrect

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