(15a) This is because neutron is very light when compared with - TopicsExpress

(15a) This is because neutron is very light when compared with alpha particle (15b) i. They are negatively charged ii. They can cause florescent iii. They can ionize gases at high tempature iv. They are fast moving electron (15c) Given, wavelength = 5.0*10^-7m = 7.0*10^-7m i. Blue light has a greater number of electrons because it has a shorter wavelenght ii. red light has higher maximum kinetic energy because it was first deflected (15di) Given Wo = 2.0eV, wavelenght= 5.0*10^-7m i. Wo = hf, Wo= 2*1.6*10^-19 =3.2*10^-19J 3.2*10^-19 = 6.6 *10^-34*Fo Fo = (3.2*10^-19 )/ (6.6*10^-34) = 0.48*10^-19+34 fo= 0.48*10^15 fo= 4.8*10^14Hz 3.0*10^8 = 4.8*10^14*wavelenght wavelenght = 3.0*10^8/ 4.8*10^14 = 0.625*10^-6 wavelenght = 6.25*10^-7 (15dii.) 1/2mV^2 = hf - hfo Kinetic energy = E-Wo E = hc/wavelenght = (6.6 * 10^-34*3.0*10^8) / 5.0*10^-7 E = 19.8 *10^-26 / 5.0*10^-7 E = 3.96 * 10-19J Kinetic energy = 3.96* 10^-19 - 3.2*10^-19J = 0.76* 10^-19J =7.6 *10^-20J (13a) (i) Object distance(u)= a Image distance(v) = b f^2 = ab (ii) 1/a+1/b = 1/f 1/16 - 1/25 = 1/f 1/f = 25-16/ 400 1/f = 9/400 f = 400/9 = 44.4cm (iii) Magnification(M) = v/u =b/a M= 25/16 = 1.5625 M=1.6 (13b) (i) Principal focus is the point mid-way between the pole of a curved mirror and the centre of curvature in which all rays that are parallel and close to the principal axis appear to converge or diverge (ii) Optical Centre: this is the centre of the lens (iii) Focal lenght: This is the distance between the pole and the principal focus of mirror or lens