The potential energy between two atoms is giv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equilibrium distance $r_0$ occurs when the potential energy $U(r)$ is at a minimum,which implies that the force $F(r) = -dU/dr = 0$.Given $U(r

Vedclass · Accepted Answer

$(2a/b)^{1/6}$

Vedclass · Answer

(A) The equilibrium distance $r_0$ occurs when the potential energy $U(r)$ is at a minimum,which implies that the force $F(r) = -dU/dr = 0$.Given $U(r) = ar^{-12} - br^{-6}$.Differentiating with respect to $r$:$dU/dr = a(-12)r^{-13} - b(-6)r^{-7} = -12ar^{-13} + 6br^{-7}$.Setting $dU/dr = 0$ for equilibrium:$-12ar_0^{-13} + 6br_0^{-7} = 0$.$6br_0^{-7} = 12ar_0^{-13}$.$b/2a = r_0^{-13} / r_0^{-7} = r_0^{-6}$.$r_0^6 = 2a/b$.Therefore,$r_0 = (2a/b)^{1/6}$.

The potential energy between two atoms is given by the function $U(r) = a/r^{12} - b/r^{6}$. Find the equilibrium distance between them.

Similar Questions

If a body of mass $200\, g$ falls from a height $200\, m$ and its total $P.E.$ is converted into $K.E.$ at the point of contact of the body with the earth's surface,then what is the decrease in $P.E.$ of the body at the contact? $(g = 10\, m/s^2)$ ............ $J$

The work done in moving a body of mass $2 \,kg$ to a height of $4 \,m$ from the surface of the earth is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,J$)

For the stability of a physical system,what should be its potential energy? Should it be more or less?

The work done in pumping out water from a cubical vessel of height $1 \ m$ is approximately ........ $J$ (Take $g = 10 \ m/s^2$ and density of water $\rho = 1000 \ kg/m^3$).

Potential energy is a......

Vedclass Test Series

Exam Paper Generator

Online Exam Module