Question

The lowest energy of an electron trapped in a rigid box is 4.19 eV. Find the width of the box in A.U. (e = 1.6 x 10-19 C, h = 6.63 x 10-34 J-sec, me = 9.1 x 10-31 kg) *​

Answer 1

Given info : The lowest energy of an electron trapped in a rigid box is 4.19 eV.

To find : The width of the box in A.U is...

solution : Lowest energy of an electron trapped in a rigid box is given by, E = h²/8mL²

where, L is width of the box, m is mass of electron and h is Plank's constant.

here, E = 4.19 eV = 4.19 × 1.6 × 10^-19 J

m = 9.1 × 10^-31 kg , h = 6.63 × 10^-34 Js

so, L² = (6.63 × 10^-34)²/(8 × 9.1 × 10^-31 × 4.19 × 1.6 × 10^-19)

= (43.9569 × 10^-68)/(488.0512 × 10^-50)

= 0.09 × 10^-18

= 9 × 10^-16 m

L = 3 × 10^-8 m

= 2.00537614 × 10^-19 AU ≈ 2 × 10^-19 AU

Therefore the width of the rigid box in AU is 2 × 10^-19 AU