[題目] fermion gas
想問一下下面這題
Prove that the average speed of a fermion gas particle at T=0K is (3/4)Vf,where
the Fermi velocity Vf is defined by εF = (1/2)mVf^2.
我的想法是 因為T=0K的average energy/per fermion 是 (3/5)εF
所以U = (3/5)εF = (1/2)mV^2
又εF = (1/2)mVf^2
故 U = (3/5)*(1/2)mVf^2 = (1/2)mV^2
所以 V(average speed) = √(3/5) Vf
可是這答案不是(3/4)Vf 但有點接近
(3/4)=0.75
√(3/5) = 0.77
想問一下要如何解才對?
感恩~~~
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◆ From: 123.195.2.59
推
12/15 01:49, , 1F
12/15 01:49, 1F
→
12/15 01:50, , 2F
12/15 01:50, 2F
εF
f(ε)代1 所以變成 ∫ v(ε)dε=? 因為v(ε)是所求的 等號右邊是等於?
0
※ 編輯: IAMROGER 來自: 123.195.2.59 (12/15 02:45)
推
12/15 04:27, , 3F
12/15 04:27, 3F
推
12/15 04:33, , 4F
12/15 04:33, 4F