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The de Broglie wavelength of an neutron corresponding to root mean square speed at `927^(@)C` is `lambda`. What will be the de Broglie wavelength of the neutron corresponding to root mean square speed at `27^(@)C`?A. `(lambda)/(2)`B. `lambda`C. `2 lambda`D. `4 lambda` |
Answer» Correct Answer - C `K.E.` of neutron `E=(3)/(2)KT` `lambda_(d)=(h)/(p)=(h)/(sqrt(2mE))=(h)/(sqrt(2mxx(3)/(2)kT))` `rArr lambda_(2)=lambdasqrt(((927+273))/(27+273))= 2 lambda` |
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