0-v-4-e-mv-2-2KT-dv-solve-it-

Question Number 27447 by raman last updated on 07/Jan/18

{\int_{0}}^{\infty} v^{4} e \frac{- {mv}^{2}}{2 KT} dv

solve it

Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18

l e t x = \frac{m}{2 K T} v^{2} v = \frac{\sqrt{2 K T}}{\sqrt{m}} \times x^{\frac{1}{2}}

d x = \frac{m}{2 K T} \times 2 v d v

= \frac{m}{K T} \times \frac{\sqrt{2 K T}}{\sqrt{m}} \times x_{}^{\frac{1}{2}} d v s o d v = \frac{K T}{m} \times \frac{\sqrt{m}}{\sqrt{2 K T}} \times x^{- \frac{1}{2}} \times

d x \frac{}{}

\int_{0}^{\infty} \frac{{(2 K T)}^{2}}{m^{2}} \times x^{2} \times e^{- x} \times \frac{\sqrt{K T}}{\sqrt{m}} \times \frac{1}{\sqrt{2}} \times x^{\frac{- 1}{2}} \times d x

\int_{0}^{\infty} \frac{2^{\frac{3}{2}} \times {(K T)}^{\frac{5}{2}}}{m^{\frac{5}{2}}} e^{- x} \times x^{\frac{3}{2}} \times d x

= {[\frac{8 {(K T)}^{5}}{m^{5}}]}^{\frac{1}{2}} \times \int_{0}^{\infty} e^{- x} \times x^{\frac{3}{2}} d x

= d i t t o \times ⌈ (\frac{5}{2}) {\int_{0}^{\infty} e^{- x} \times x^{n - 1} = ⌈ (n)}

n o w ⌈ (\frac{5}{2}) = \frac{3}{2} \times \frac{1}{2} \times \sqrt{Π}

= {\frac{8 {(K T)}^{5}}{m^{5}}}^{\frac{1}{2}} \times \frac{3}{4} \times \sqrt{Π}

p l s c h e c k