Question

What is the energy difference ΔE between the l=0 and l=1 rotational states of diatomic hydrogen? Use 1.05×10−34J⋅s for ℏ. The equilibrium separation of the two hydrogen atoms in diatomic hydrogen is 0.074nm. The mass of a single hydrogen atom is 1.67×10−27kg.

Answer 1

Answer:

$2.4\times 10^{-21} J$

Explanation:

We are given that

l=0 and l=1

$h=1.05\times 10^{-34} J\cdot s$

$r=0.074nm=0.074\times 10^{-9} m$

$1nm=10^{-9} m$

Mass of single hydrogen,m=$1.67\times 10^{-27} kg$

$E_l=\frac{l(l+1)h^2}{mr^2}$

Energy difference,$\Delta E=E_{l=1}-E_{l=0}$

$\Delta E=\frac{h^2}{mr^2}(1(1+1))-\frac{h^2}{mr^2}(0(0+1))=\frac{h^2}{mr^2}(1(1+1))=\frac{2h^2}{mr^2}$

$\Delta E=\frac{2\times (1.05\times 10^{-34})^2}{1.67\times 10^{-27}\times (0.074\times 10^{-9})^2}$

$\Delta E=2.4\times 10^{-21} J$