Exercise 2:
En = n2h2/(8mL2)
L = nh*(8mEn)-1/2
= 3*6.63*10-34Nms*(8*9.11*10-31kg*4.7*1.6*10-19Nm)-1/2
= 8.50*10-10 m
Exercise 9:
Form the first derivate of the probability density P(x), dP(x)/dx, set it to zero.
è maxima and minima
Use the second derivative to distinguish between maxima and
minima. This will give a condition x=Nl/n for odd n
and x=Nl/2n for even n.
Exercise 13: similar to sample problem 47-5, page
1065
1) hf=E3-E2
f = (E3-E2)/h
= -me4(8ε02h3)-1(1/32
– 1/22) =
(9.1094*10-31kg)(1.6022*10-19C)4(8*(8.8542*10-12F/m)2(6.6261*10-34Js)3(1/32
– 1/22) = 4.5695*1014Hz
λ=c/f =
2.9979*108m/s / 4.5695*1014Hz = 656.1 nm
2) hf=E4-E2
f = (E4-E2)/h
= -me4(8ε02h3)-1(1/42
– 1/22) =
(9.1094*10-31kg)(1.6022*10-19C)4(8*(8.8542*10-12F/m)2(6.6261*10-34Js)3(1/42
– 1/22) = 6.1688*1014Hz
λ=c/f = 2.9979*108m/s / 6.1688*1014Hz = 486.0 nm
3) hf=E5-E2
f = (E5-E2)/h
= -me4(8ε02h3)-1(1/52
– 1/22) =
(9.1094*10-31kg)(1.6022*10-19C)4(8*(8.8542*10-12F/m)2(6.6261*10-34Js)3(1/52
– 1/22) = 6.9090*1014Hz
λ=c/f = 2.9979*108m/s / 6.9090*1014Hz = 433.9 nm
4) hf=E6-E2
f = (E6-E2)/h
= -me4(8ε02h3)-1(1/62
– 1/22) =
(9.1094*10-31kg)(1.6022*10-19C)4(8*(8.8542*10-12F/m)2(6.6261*10-34Js)3(1/62
– 1/22) = 7.3111*1014Hz
λ=c/f = 2.9979*108m/s / 7.3111*1014Hz = 410.0 nm
5) hf=E7-E2
f = (E7-E2)/h
= -me4(8ε02h3)-1(1/72
– 1/22) =
(9.1094*10-31kg)(1.6022*10-19C)4(8*(8.8542*10-12F/m)2(6.6261*10-34Js)3(1/72
– 1/22) = 7.5536*1014Hz
λ=c/f
= 2.9979*108m/s / 7.5536*1014Hz = 396.9 nm
Problem 1:
E111=h28-1m-1L-2(n12+n22+n32)=(6.63*10-34Js)2*0.125*(9.1094*10-31kg)-1(250*10-9m)-2(12+12+12) = 2.90*10-24J = 18.1 μeV
E112=E121=E211=E111(12+12+12)-1(12+12+22) = 36.1 μeV
E221=E212=E122=E111(12+12+12)-1(12+22+22)
= 54.2 μeV
E113=E131=E311=E111(12+12+12)-1(12+12+32)
= 66.3 μeV
E222=E111(12+12+12)-1(22+22+22)
= 72.3 μeV