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Question: The kinetic energy of a neutron beam is 0.0837 *eV*. The half-life of neutrons is 693*s* and the mas...

The kinetic energy of a neutron beam is 0.0837 eV. The half-life of neutrons is 693s and the mass of neutrons is 1.675×10−271.675 \times 10^{- 27}kg. The fraction of decay in travelling a distance of 40m will be

A

10−310^{- 3}

B

10−410^{- 4}

C

10−510^{- 5}

D

10−610^{- 6}

Answer

10−510^{- 5}

Explanation

Solution

v=2Em=2×0.0837×1.6×10−191.675×10−27v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 0.0837 \times 1.6 \times 10^{- 19}}{1.675 \times 10^{- 27}}}= 4 × 103 m/sec

∴ Time taken by neutrons to travel a distance of 40 m Δt′=404×103=10−2sec⁡\Delta t' = \frac{40}{4 \times 10^{3}} = 10^{- 2}\sec

∴ dNdt=λN⇒dNN=λdt\frac{dN}{dt} = \lambda N \Rightarrow \frac{dN}{N} = \lambda dt

∴ Fraction of neutrons decayed in ∆t sec in

ΔNN=λΔt=0.693TΔt=0.693693×10−2=10−5\frac{\Delta N}{N} = \lambda\Delta t = \frac{0.693}{T}\Delta t = \frac{0.693}{693} \times 10^{- 2} = 10^{- 5}