Potential energy of a diatomic molecule is given by U = frac{A}{r^{12}} - frac{B}{r^6} . A and

English

Gujarati

Hindi

11.Thermodynamics

normal

The potential energy of a diatomic molecule is given by $U$ = $\frac{A}{r^{12}} - \frac{B}{r^6}$.$A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is

A

$(\frac{A}{B})^\frac{1}{6}$

B

$(\frac{2A}{B})^\frac{1}{6}$

C

$(\frac{A}{2B})^\frac{1}{6}$

D

None of these

Solution

At equilibrium $\frac{\mathrm{d} \mathrm{U}}{\mathrm{dr}}=0$

$\frac{-12 A}{r^{13}}+\frac{6 B}{r^{7}}=0$

$r=\left(\frac{2 A}{B}\right)^{1 / 6}$

Standard 11

Physics

Share

Similar Questions

Calculate heat given to gas during process $ABA$ in figure …… $J$

normal

Initial pressure and volume of a gas are $P$ and $V$ respectively. First it is expanded isothermally to volume $4\ V$ and then compressed adiabatically to volume $V.$ The final pressure of gas will be (given $\gamma = 3/2)$

normal

One mole of an ideal diatomic gas undergoes a transition from $A$ to $B$ along a path $AB$ as shown in the figure

The change in internal energy of the gas during the transition is

normal

Two rigid boxes containing different ideal gases are placed on a table. Box $A$ contains one mole of nitrogen at temperature $T_0$ , while box $B$ contains one mole of helium at temperature $(7/3)\, T_0$ . The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes), then the final temperature of gases $T_f$ , in terms of $T_0$ is

normal

$P-V$ diagram of $2\, g$ of $He$ gas for $A \to B$ process is shown. What is the heat given to the gas ?

normal