heyy guyz..help me plss
An electron and a proton, each has a de-broglie wavelength of 1.6 nm
1- find the ratio of their momenta.
2- compare the kinetic energy of the proton with that of the electron.
#class 12
Answers
Answered by
0
Answer:
Answer
(i)
λ=
p
e
h
and λ
p
=
p
p
h
,λ
e
=λ
p
=1.00nm
So,
λ
p
λ
e
=
p
e
p
p
=
1
1
⇒
p
e
p
p
=
1
1
=1:1
(ii)
From relation K=
2
1
mv
2
=
2m
p
2
K
e
=
2m
e
p
e
2
and K
p
=
2m
p
p
p
2
K
e
K
p
=
2m
p
p
p
2
×
p
e
2
2m
e
=
m
p
m
e
Since m
e
<<<m
p
=
1.67×10
−27
9.1×10
−31
=5.4×10
−4
Similar questions