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The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and mass `m`, isA. `(h)/(sqrt(mkT))`B. `(h)/(sqrt(3mkT))`C. `(2h)/(sqrt(3mkT))`D. `(2h)/(sqrt(mkT))` |
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Answer» Correct Answer - B Thinking process de Broglie wavelengfth associated with a moving particle can be given as `lambda=h/pr=(h)/(sqrt(2m(KE))` At thermal equilibrium, temperature of neutron and heavey water will be same . This commo n temp[erature is given as T. Also, we know that, kinetic energy of a particle `KE=(p^(2))/(2m)` where,P = momentum of the particle m=mass of the particle Kinetic energy of the neutron is `KE=(3)/(2)KT` `therefore` de-Broglie wavelength of the neutron `lambda=h/p=(h)/(sqrt(2m(KE))=(h)/(sqrt(2mxx(3)/(2)kT))=(h)/(sqrt(3mKT))` |
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