# Show that the function has a maximum at E = ((b*k*T)/2)^2/3?

One way of writing the Gamow Peak function is f(E)=e^(-(E/(k∗T))−(b/√E)) . Show that the function has a maximum at E=((b∗k∗T)/2)^2/3 .

### 1 Answer

- DouglasLv 77 years agoBest Answer
Given:

f(E)=e^(-(E/(k∗T))−(b/√E)) = {e^(-(E/(k∗T)))}{e^−(b/√E)}

Take the derivative using the product rule:

Let u(E) = {e^(-(E/(k∗T)))}, then u'(E) = {-(1/(k∗T)e^(-(E/(k∗T)))}

Let v(E) = {e^−(b/√E)}, then v'(E) = {-b(-1/2)(E^(-3/2))e^−(b/√E)}

d{f(E)}/dE = u'(E)v(E) + u(E)v'(E)

d{f(E)}/dE = {-(1/(k∗T)e^(-(E/(k∗T)))}{e^−(b/√E)} + {e^(-(E/(k∗T)))}{-b(-1/2)(E^(-3/2))e^−(b/√E)}

Set the right side equal to zero:

{-(1/(k∗T)e^(-(E/(k∗T)))}{e^−(b/√E)} + {e^(-(E/(k∗T)))}{-b(-1/2)(E^(-3/2))e^−(b/√E)} = 0

Common factors e^(-(E/(k∗T))) and {e^−(b/√E)} can be divided out of the equation:

{-(1/(k∗T)} + {-b(-1/2)(E^(-3/2))} = 0

Multiply by (k∗T):

-1 + -(k∗T ∗ b)(-1/2)(E^(-3/2)) = 0

-1 + (k∗T ∗ b)/2)(E^(-3/2)) = 0

Add 1 to both sides:

((k∗T ∗ b)/2)(E^(-3/2)) = 1

Multiply by E^(3/2) and flip the equation:

E^(3/2) = (k∗T ∗ b)/2

Raise to the 2/3 power:

E = ((k∗T ∗ b)/2)^2/3