Question

Outside the nucleus, the neutron itself is radioactive and decays into a proton, an electron, and an antineutrino. The half-life of a neutron (mass = 1.675 × 10-27 kg) outside the nucleus is 10.4 min. On average, over what distance x would a beam of 5.55-eV neutrons travel before the number of neutrons decreased to 75.0% of its initial value? Ignore relativistic effects.

Answer 1

value of decay constant

= 0.693/t_1/2

= 0.693/10.4*60

= 0.00111

and

N(t) = No*e^-t

0.75 No = No *e^-t

log(0.75) = -*t

t = 258.55 s

and

E = 5.55 ev = 5.55/6.24*10¹⁸ J

E = 0.889*10^-18 J

this energy is equal to the kinetic energy of the neutron

1/2*m*v² = 0.889*10^-18

v = sqrt(2*0.889*10^-18 /1.67*10^-27)

v = 32629.29 m/s

Now,

x = v*t

x = 32629.29*258.55

x = 8.43*10⁶ m

distance x travel before the number of neutrons decreased to 75.0% of its initial value is 8.43*10⁶ m.