Question Number 158694 by ilhamdiii last updated on 07/Nov/21
$${Help}\:{me}\:{sir} \\ $$$$\: \\ $$$${in}\:{phase}\:{sppace}\:\:{d}^{\mathrm{3}} {p}={dp}_{{x}} {dp}_{{y}\:} {dp}_{{z}} \:\:{then} \\ $$$${find} \\ $$$$\:\int_{\mathrm{0}} ^{\infty} \:{P}^{\mathrm{2}\:} {e}^{\frac{{p}^{\mathrm{2}} }{\mathrm{2}{mKT}\:}} \:\:{d}^{\mathrm{3}} {p}\:\:=…. \\ $$$$ \\ $$$$ \\ $$
Answered by MrGaster last updated on 06/Jan/25
$$\mathrm{Let}\:{x}=\frac{{p}^{\mathrm{2}} }{\mathrm{1}{m}}\Rightarrow{p}=\sqrt{\mathrm{2}{mKT}}×{x}^{\frac{\mathrm{1}}{\mathrm{2}}} {dp}=\sqrt{\frac{{m}}{\mathrm{2}{Kp}}}×{x}^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} {p}^{\mathrm{2}} {e}^{−{p}^{\mathrm{2}} /\left(\mathrm{2}{mKT}\right)} {d}^{\mathrm{3}} {p} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\mathrm{2}{mkT}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} ×{x}^{\frac{\mathrm{3}}{\mathrm{2}}} ×{e}^{−{x}} ×\left(\frac{\sqrt{{m}}}{\:\sqrt{\mathrm{2}{KT}}}\right)×\frac{\mathrm{1}}{\:\sqrt{{x}}}{dx} \\ $$$$=\frac{\left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} ×\left({mKT}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\:}{\:\sqrt{\mathrm{2}\pi}}\int_{\mathrm{0}} ^{\infty} {x}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{1}/\mathrm{2}} {e}^{−{x}} {dx}=\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\frac{\left(\mathrm{2}^{\mathrm{3}/\mathrm{2}} ×\left({mKT}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right)}{\mathrm{2}\sqrt{\mathrm{2}\pi}}×\sqrt{\pi} \\ $$$$=\frac{\sqrt{\mathrm{2}}×\left({mKT}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}\sqrt{\mathrm{2}\pi}}×\sqrt{\pi} \\ $$$$=\frac{\sqrt{\mathrm{2}}×\left({mKT}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\:\sqrt{\pi}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {p}^{\mathrm{2}} {e}^{−{p}^{\mathrm{2}} /\left(\mathrm{2}{mKT}\right)} {d}^{\mathrm{3}} {p}=\frac{\sqrt{\mathrm{2}}×\left({mKT}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\:\sqrt{\pi}} \\ $$