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Question Number 158694 by ilhamdiii last updated on 07/Nov/21

H e l p m e s i r

i n p h a s e s p p a c e d^{3} p = {d p}_{x} {d p}_{y} {d p}_{z} t h e n

f i n d

\int_{0}^{\infty} P^{2} e^{\frac{p^{2}}{2 m K T}} d^{3} p = \dots .

Answered by MrGaster last updated on 06/Jan/25

Let x = \frac{p^{2}}{1 m} \Rightarrow p = \sqrt{2 m K T} \times x^{\frac{1}{2}} d p = \sqrt{\frac{m}{2 K p}} \times x^{- 1 / 2} d x

\int_{0}^{\infty} p^{2} e^{- p^{2} / (2 m K T)} d^{3} p

= \int_{0}^{\infty} {(2 m k T)}^{\frac{3}{2}} \times x^{\frac{3}{2}} \times e^{- x} \times (\frac{\sqrt{m}}{\sqrt{2 K T}}) \times \frac{1}{\sqrt{x}} d x

= \frac{(2^{\frac{3}{2}} \times {(m K T)}^{\frac{3}{2}})}{\sqrt{2 π}} \int_{0}^{\infty} x^{\frac{1}{2}} e^{- x} d x

\int_{0}^{\infty} x^{1 / 2} e^{- x} d x = Γ (\frac{3}{2}) = \frac{\sqrt{π}}{2}

\frac{(2^{3 / 2} \times {(m K T)}^{\frac{3}{2}})}{2 \sqrt{2 π}} \times \sqrt{π}

= \frac{\sqrt{2} \times {(m K T)}^{\frac{3}{2}}}{2 \sqrt{2 π}} \times \sqrt{π}

= \frac{\sqrt{2} \times {(m K T)}^{\frac{3}{2}}}{\sqrt{π}}

\int_{0}^{\infty} p^{2} e^{- p^{2} / (2 m K T)} d^{3} p = \frac{\sqrt{2} \times {(m K T)}^{\frac{3}{2}}}{\sqrt{π}}

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