Question

3 attempts left Check my work Calculate the energy of the four lowest energy levels of an electron in a one-dimensional box with a length of 1.30 nm. For n-1: x 10 J For -2: 10J For n-3: For n-4: X 10

Answer 1

Energy of electron in one dimensional box is E=n2h2/8mL2

Solution Energy of electron in a cone-dimensional box is E = n2h² 8m L2 where. n energy level. hs Plankis Constant. = 6163x10-34 yos mt Mass of an electron. = 9. 1X20-3) Lt Length of one-dimensional box. Given that L = 1.30 nm. Convert nm tom Inm = 10-9 m L=1.30 X10-9 m So for n=1 . E= no 8m12 E = (1)’x(6:63 X 16-3472 8x9.1.X20-31x (130 X 10.9j2 E = 0.35728 X1079 9 ogja ya E = 3,57 x 10-20 % /

(i) For n=2 E = (23x (6.63 X 70-3472 8x 9,1X10-31x (1.30 X10-9j2 109j2 y E = 4x66-63)? x 10-68 8x911 X 2.30X1:30 X 70-34 X10-18 v E = 175.8276 X10+687 1231032 X 10-49 TE= 1.429 X10-19 11 (iii) For n=3 E = (352x66-63x10-342 8x9-1X10-31 x (130 X119)2 E = 9x6.63 x6:63 x 10-68 8x9.1X 130X1-30 X1031x10-18 E = 395.6121X10684 123,032 X10-49 E = 3:21 X 10-19 j]

Solution (iv) n=4 E = (412x (6+63 X 10-34)2 88 9.18 10-3'x (1130X10-9j2 E = 16 X6163x6.63 x 10-68 8x9.1X1:30X1:30 X10-31 x 10-18 E = 703.3104 x 10-68 123.032x10-49 E = 5:716X10-19 ja