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

asking for x in meters

Answer 1

Using a formula, we have

N / N₀ = (1/2)^t/t_1/2

We know that, the number of neutrons decreased to 75% of its initial value.

(0.75 N₀) / N₀ = (1/2)^t/(10.4)

(0.75) = (1/2)^t/(10.4)

ln (1.5) = t / (10.4 min)

t = 4.21 min

convert min into sec -

t = 252.6 sec

Using a formula, we have

v = c 1 - E₀² / E²

where, c = speed of light = 3 x 10⁸ m/s

E₀ = rest energy = m₀ c²

v = (3 x 10⁸ m/s) 1 - [(1.675 x 10^-27 kg) (3 x 10⁸ m/s) / (7.5302 x 10^-19 J)]²

v = (3 x 10⁸ m/s) (0.9949)

v = 2.98 x 10⁸ m/s

Therefore, we get

distance x = v t = (2.98 x 10⁸ m/s) (252.6 s)

x = 7.52 x 10¹⁰ m