Question

In free space the intensity of 5 eV neutron beam is reduced by a factor of one half. Half life is t_1/2 = 12.8 min. The distance travelled by neutron beam is-

• A. 2800 km
• B. 23800 km
• C. 28 km
• D. 2 km

Answer 1

Answer: (B)

23800 km

Answer 2

Speed of the neutrons in beam is

= K = 5eV

v = $\sqrt { \frac { 2 ( 5 ) \times 1.6 \times 10 ^ { - 19 } } { 1.67 \times 10 ^ { - 27 } } }$

v = 31 km/sec

During a time of t_1/2 = 12.8 min, half the neutrons will have decayed from the beam. The distance travelled by the undecayed during this time is d = nt = (31 km/s) (12.8 min) (60 s/min)

= 23800 km