JEE Advanced Model Paper 2 with solutions for online practice

velocity of beam = √

2E

m

=
= √2×0.05×108 = √13×106 = 4 × 103ms−1
Time required to travel a distance of 4 km =

distance

seed

=

4×1000

4×102

= 1 s
Now, decay rate

dN

dt

= λN ⇒

dN

N

= λ (Δt)
Fraction of neutrons that decay in 1 s
=

dN

N

=

ΔN

N

= λΔt = λ =

log‌2

T1∕2

=

0.0693

693

= 10−3
⇒

N2−N1

N

= 10−3
⇒ N2−N1 = 10−3 × N = 10−3 × 104 = 10
So, 10 neutrons will decay