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 4.48-eV neutrons travel before the number of neutrons decreased to 75.0% of its initial value? Ignore relativistic effects.

Answer 1

energy of neutrons = 4.48ev

mass of neutron = 1.675 × 10-27 kg

half life = 10.4 minutes

Time required for the netrons to decay to 75% of its initial value.

N is the number of neutrons present at instant time 't'

No is the number of neutrons present at time t=0 ( intial concentration)

is the decay constant

10.4 minutes = 624 seconds

= 0.0011/second

ln (0.75)= -.t

t = 261.5 seconds

K.E of neutron    = 4.48ev= 4.48 * 1.6* joules

velocity =

velocity = 2.92*

distance travelled =velocity * time =  2.92** 261.5 seconds= 763.5*

distance travelled =