Question

Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly :

(Take $\frac{\pi}{2}$= 12.8 min)

• A. 6000 km
• B. 12000 km
• C. 18000 km
• D. 24000 km

Answer 1

Answer: (D)

24000 km

Answer 2

$\frac { 1 } { 2 }$mv² = K

$\frac { 1 } { 2 }$ (1.67 × 10^–27) v² = 5 × 1.6 × 10^–19

̃ v = 31 km/s

d = vt = 31 × 10³ × 12.8 × 60 I 24000 km