Question

Alpha particles emitted from radioactive radium has an energy of 4.8 MeV (million electron volts). Calculate the de Broglie wavelength of one of these alpha particles, given the following data: (mass of alpha particle = 6.6 x 10-24 g h=6.6 x 10-27 erg sec 1.0 MeV = 1.6 x 10-6 erg 1 erg=1 g cm² sec-2)

Answer 1

0 solution Radium is radioactive element that erits alpha particles Chelium nuclei, Helt), beta Particles (electrons) and gamma rays. Energy depends on frequency or wavelength according to the relation R E = hr= Che/1) - 0 From Einstein's formula E = mc2 Where m = the mass of the particle and c is the Speed of light; the particle's momentum is po me !: p= me = ( Elc) - ☺ Substituting (1) into eq 5 yields but Ching is also equal to my hence P= (h/2) = mv - Equestion (4) is the de Broglie equation. Rearranging equation (4) yields 2 = (h/mv) - 5 The kinetic energy, KE of the alpha particularos is given as 4.8 Mev. that is, KE = 4.8 Mev. But, KE = (1/2) Mia :(1/2) mv2 = 4.8 Mev.

and V? = 52(48 MeV)7)m] = [TC9.6 Mev)*1.6 xio erg), CI MeV} {(19 cm² sec 2) (1 erg)}}) (6.6x1024g)] = 2.32x1068 ccm/sec and r={{2.32x 108 ccm/sec? I = 1.52x109 ccm) See) This is the velocity of the alpha particles. Substituting this velocity into equestion 6, where we have 2= {{(6.6*1027 erg sec) {(lq cm sec 2)/(1 erg)]} IEC616x1074)[1.52 x lo'ccm/sec)}}] or 2 = 606 x 10'3cm