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 :)
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 :)
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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:
(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