Kichah Corr

‫النموذجى‬ ‫التصحيح‬ ‫الول‬ ‫التمرين‬ 1‫ـ‬J 2 – 10 = c E 2 Vm 2 /1 = c E s/ m 0.45 = V 2‫ـ‬A E = B E 0h g m = J 2 – 10 m 2 – 10 = 0 h 3‫ـ‬ 1c E 2pp E 2c E 1pp E 4‫ـ‬ α 0h 0 h α s o c l - l = 0 h =α s o c 0 8.1=α ‫الثانى‬ ‫التمرين‬ F 1‫ـ‬β R P f 2‫ـ‬0=R W J3... More

‫النموذجى‬ ‫التصحيح‬ ‫الول‬ ‫التمرين‬ 1‫ـ‬J 2 – 10 = c E 2 Vm 2 /1 = c E s/ m 0.45 = V 2‫ـ‬A E = B E 0h g m = J 2 – 10 m 2 – 10 = 0 h 3‫ـ‬ 1c E 2pp E 2c E 1pp E 4‫ـ‬ α 0h 0 h α s o c l - l = 0 h =α s o c 0 8.1=α ‫الثانى‬ ‫التمرين‬ F 1‫ـ‬β R P f 2‫ـ‬0=R W J3 10 .721.25 = F W 0. 5 0. 5 0. 5 0. 5 0. 5 0. 5 2 0. 5 0. 5 0. 5 0. 5 J3 10 .361.25 - = PW J f 850 - = fW 3‫ـ‬0=)F ) W Σ 3 10.721.25+3 10.361.25-f 850 - = 0 N 423.52 = f 4‫ـ‬ 5‫ـ‬2pp E 1pp E ‫الثالث‬ ‫التمرين‬ 1‫عند‬ ‫ـ‬A‫الطاقة‬ ‫شكل‬pp E ‫عند‬B‫الطاقة‬ ‫شكل‬pp E+C E ‫عند‬C‫الطاقة‬ ‫شكل‬C E ‫عند‬D‫الطاقة‬ ‫شكل‬ep E 2‫ـ‬ 1pp E 2e pE 2pp E 1e pE 3‫ـ‬D E = A E J 5 = 2 X K 2 /1 = h g m 0.5 0. 5 0. 5 0. 5 1 1 1 1 1 1 1 0.5 0.5 0h - l l Less