- #1
Victoire
- 3
- 0
Hi, I'm stuck on a problem. I have this function that represents the velocity:
f(v)=v2e(-mv2)/(2kT)
T is the temperature and equals 293.15K
m = 4.65 .10-26 kg
k = 1.38 .10-23 J/K
The problem wants us to find the derivative of the function. I found:
df/dv= (2v) e(-mv2)/(2kT)+v2e(-mv2)/(2kT)(-2mv/2kT)
= ve(-mv2)/(2kT) (2- (mv2/kT))
I know the derivative I found is right, but now, I have to find the velocity for df/dv=0
We know v>0 and ex>0
So, df/dv=0 => (mv2)/kT = 2
=> v= √(2kT/m)
That's where I'm stuck. The book tells me the answer is 1500km/h.
I don't understand this solution. I calculated many many times and didn't found the right solution. Can someone please help me?
f(v)=v2e(-mv2)/(2kT)
T is the temperature and equals 293.15K
m = 4.65 .10-26 kg
k = 1.38 .10-23 J/K
The problem wants us to find the derivative of the function. I found:
df/dv= (2v) e(-mv2)/(2kT)+v2e(-mv2)/(2kT)(-2mv/2kT)
= ve(-mv2)/(2kT) (2- (mv2/kT))
I know the derivative I found is right, but now, I have to find the velocity for df/dv=0
We know v>0 and ex>0
So, df/dv=0 => (mv2)/kT = 2
=> v= √(2kT/m)
That's where I'm stuck. The book tells me the answer is 1500km/h.
I don't understand this solution. I calculated many many times and didn't found the right solution. Can someone please help me?