JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 23)
The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as
$$U = {\alpha \over {{r^{10}}}} - {\beta \over {{r^5}}} - 3$$
where, $$\alpha$$ and $$\beta$$ are positive constants. The equilibrium distance between two atoms will be $${\left( {{{2\alpha } \over \beta }} \right)^{{a \over b}}}$$, where a = ___________.
$$U = {\alpha \over {{r^{10}}}} - {\beta \over {{r^5}}} - 3$$
where, $$\alpha$$ and $$\beta$$ are positive constants. The equilibrium distance between two atoms will be $${\left( {{{2\alpha } \over \beta }} \right)^{{a \over b}}}$$, where a = ___________.
Answer
1
Explanation
$$F = - {{dU} \over {dr}}$$
$$F = - \left[ { - {{10\alpha } \over {{r^{11}}}} + {{5\beta } \over {{r^6}}}} \right]$$
for equilibrium, F = 0
$${{10\alpha } \over {{r^{11}}}} = {{5\beta } \over {{r^6}}}$$
$${{2\alpha } \over \beta } = {r^5}$$
$$r = {\left( {{{2\alpha } \over \beta }} \right)^{1/5}}$$
$$ \therefore $$ $$a = 1$$
$$F = - \left[ { - {{10\alpha } \over {{r^{11}}}} + {{5\beta } \over {{r^6}}}} \right]$$
for equilibrium, F = 0
$${{10\alpha } \over {{r^{11}}}} = {{5\beta } \over {{r^6}}}$$
$${{2\alpha } \over \beta } = {r^5}$$
$$r = {\left( {{{2\alpha } \over \beta }} \right)^{1/5}}$$
$$ \therefore $$ $$a = 1$$
Comments (0)
