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

`2.4* 10^(-21) J`

Step-by-step explanation:

We are given that

l=0 and l=1

`h=1.05* 10^(-34) J\cdot s`

`r=0.074nm=0.074* 10^(-9) m`

`1nm=10^(-9) m`

Mass of single hydrogen,m=
`1.67* 10^(-27) kg`

`E_l=(l(l+1)h^2)/(mr^2)`

Energy difference,
`\Delta E=E_(l=1)-E_(l=0)`

`\Delta E=(h^2)/(mr^2)(1(1+1))-(h^2)/(mr^2)(0(0+1))=(h^2)/(mr^2)(1(1+1))=(2h^2)/(mr^2)`

`\Delta E=(2* (1.05* 10^(-34))^2)/(1.67* 10^(-27)* (0.074* 10^(-9))^2)`

`\Delta E=2.4* 10^(-21) J`