Compute the energy separation between the ground and third excited states for an electron in a one-dimensional box that is 5.60 angstroms in length. express the energy difference in kj⋅mol−1.

35.51 × 10^-17 KJ/mol

Explanation:

We can calculate this using the Planck's equation:

E = hc/λ

Where E is the energy separation

h is the planck's constant with a value 6.63 × 10^-34 m^2kg/s

c is the speed of light with value 3 × 10^8m/s

λ is the wavelength with value 5.60Å. 1 Å = 10^-10 m. Hence , 5.60Å is 0.56 × 10^-9nm.

We plug the values in the equation:

E = (6.63 × 10^-34 × 3 x 10^8)/ 0.56 × 10^-9 = 35.51 × 10^-17KJ

for Planck's constant (h) and 3 × 108 m/s for the speed of light (c)

1 Å = 10-10 m

λ=Ehc

50.23 i believe is correct but don't quote me on that.