Consider the potential energy landscape shown in the graph below, U = U(x). The
ID: 1997843 • Letter: C
Question
Consider the potential energy landscape shown in the graph below, U = U(x). The x-scale is in centimeters (i.e., one centimeter per division on the graph paper), the U-scale is in Joules (i.e., one Joule per division on the graph paper). For example, the point A is at (1cm, 14 Joules), the point B is at (6cm, 1 Joule), and so on. WHICH LETTERED POINT(S) on the graph represent stable equilibrium? Why? WHICH LETTERED POINT(S) on the graph represent unstable equilibrium? Why? WHICH LETTERED POINT on the graph represents a global minimum? AT WHICH LETTERED POINT on the graph is there a negatively-directed force? Give a numerical estimate of its value in Newtons, and explain your reasoning (perhaps graphically). WHICH LETTERED POINT on the graph represents the maximum positively-directed force? Give a numerical estimate of its value in Newtons, and explain your reasoning (perhaps graphically) Suppose there is a particle bound in this potential energy landscape at point E (16cm, 3 Joules). What is the activation energy needed to liberate this particle to point G (and thus unbound from the landscape)?Explanation / Answer
(i) Points B,E are stable Equilibrium Points as if a particle placed there is displaced towards right,it will experience Force towards left.
(ii) Points D,F are at Unstable Equilibrium as The force tries to displace the particle away from Equilibrium point.
(iii) Point B has Global minima.To find one out,Find all local minima's and then look which one is least.
(iv) Force is negatively Directed at Stable Eq. points i.e B,E.And its value is found out by : F = -(dU/dx),where U is the potential Energy .
(v) Force is positively directed at Unstable Eq. points i.e at D,F.
(vi) Activation Energy = Ea = Potn. energy at F - Potn. Energy at E = 6-3 =3Joules.At point F(unstable Eq.) the particle will go to G with a slight displacement towards right.
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