Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly: (Take half-life of the neutron =12.8 min )

Question

Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly: (Take half-life of the neutron =12.8 min ) (A) 6000km (B) 12000km (C) 18000km (D) 24000km

Tardigrade Admin · Accepted Answer

The kinetic energy of the neutron beam 5 eV=(1/2) textmv2 (1/2)(1.67 × (10)- 27)v2=5× 1.6× (10)- 19 ⇒ v=31 kms- 1 distance travelled in one half-life d=vt=31× 103× 12.8× 60m∼ eq24000 km

Q. Distance in free space at which intensity of $5 e V$ neutron beam reduces to half will be nearly: (Take half-life of the neutron $= 12.8 min$ )

$6000 km$

$12000 km$

$18000 km$

$24000 km$

The kinetic energy of the neutron beam $5 e V = \frac{1}{2} mv^{2}$
$\frac{1}{2} (1.67 \times (10)^{- 27}) v^{2} = 5 \times 1.6 \times (10)^{- 19}$
$\Rightarrow v = 31$ $km s^{- 1}$
distance travelled in one half-life $d = v t = 31 \times 1 0^{3} \times 12.8 \times 60 m \sim e q 24000 km$

Q. Distance in free space at which intensity of 5eV neutron beam reduces to half will be nearly: (Take half-life of the neutron =12.8min )

6000km

12000km

18000km

24000km