Question

Calculate the energy gap in J, kJ/mol, eV, and wavenumber (cm^-1) between the level n = 5 and n = 1 of an electron confined to a microscopic 1-D box of length 2.00 Angstroms.

  1- J
  2- kJ/mol
  3 -eV
  4- cm^-1

Answer 1

ΔE = E ( 5 ) – E ( 1 )

E ( n ) = n ² h ² / [ 8 m L ² ] … where n = 1 , 2 , 3 , ⋯ , or any integer
… m = mass of the particle inside the box (the electron in this case

h is Planck’s constant; 6.63 × 10 ⁻ ³⁴ J • s

mass of the electron m = 9.11 × 10 ⁻ ³¹ kg

L = length of the box; 2.00 Angstroms or 2.0 × 10^-10 meters

ΔE = E ( 5 ) – E ( 1 )

ΔE = 5 ² h ² / [ 8 m L ² ] - 1 ² h ² / [ 8 m L ² ]

ΔE = (6.63 × 10 ⁻ ³⁴ J • s) ² / [ 8 *9.11 × 10 ⁻ ³¹ kg (2.0 × 10^-10 meters) ² ] (25-1)

ΔE =3.7*10^-17 J

energy of an electron = 3.7*10^-17 J or 3.7*10^-20 KJ

energy of e1mol electron = 6.022 × 10^{23 *}3.7*10^-20 KJ = 22281.4 KJ / mol

energy of an electron = 3.7*10^-17 J or 3.7*10^-20 KJ

1.00 J = 6.24 eV

3.7*10^-17 J * 6.24 eV / 1.00 J =23.088*10^-17 eV

wave number=k=1/λ
k = 1/λ= 2517 (1/m)

and
E=h*c*k

now calculate the wave number as follows:

3.7*10^-17 J =6.63 × 10 ⁻ ³⁴ J • s *3.00*10^8 m/s*k

k= 3.7*10^-17 J / 6.63 × 10 ⁻ ³⁴ J • s *3.00*10^8 m/s

k= 0.186*10^9 / m or 1.86*10^10 / cm