Question

A neutron travels a distance of 12m in time interval of 3.6*10^-4 s . Assuming its speed was constant , its kinetic energy is (take 1.7*10^-27 kg as the mass of neutron )

A.3.1 ev
B.4.7 ev
C.5.78 ev
D.6.91 ev
E.7.81 ev

Need soultion w/working as well :)

Answer 1

speed of neutron = distance travelled by neutron / time taken by neutron for covered that distance .

speed of neutron = 12/3.6 × 10^-4 m/s

speed of neutron ( V)= 3.3 × 10⁴ m/s

now ,
kinetic energy = 1/2mv²

K.E = 1/2 × 1.7 × 10^-27 ×( 3.3 × 10⁴)²j
= 1/2 × 1.7 × (3.3)² × 10^-19 j
=9.2565 × 10^-19 j

we know ,
1ev = 1.6 × 10^-19 j

so,
K.E = 9.2565 × 10^-19/1.6 × 10^-19 ev

= 9.2565/1.6 ev

= 5.785 ev

hence ,option ( C) is correct answer .

Answer 2

Answer:

(a) The wavelength associated with neutron, λ=1.4×10

−10

m

Mass of neutron, m=1.66×10

−27

kg

The kinetic energy is given byK=

2

1

mv

2

For de broglie wavelength and velocity,

λ=

mv

h

So,

K=

2λ

2

m

h

2

=6.75×10

21

J=4.219×10

2

eV

(b) The temperature of surrounding T=300K, m is the mass of neutron.

Boltzmann constant, K=1.38×10

−23

kgm

2

s

−2

K

−1

Average K.E. of neutron is,

K

′

=

2

3

KT

⇒

2

3

(1.38×10

−23

×300)

=6.21×10

−21

J

From de broglie equation,

λ

′

=

2K

′

m

h

⇒

2×6.21×10

−21

×1.66×10

−27

6.6×10

−34

=0.146