Can anyone tell me the from which source I can get values of parameter of closest approach for various molecules and ions, I need to use it in extended PDH formula?
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Answered by
0
Answer:
The distance of closest approach of an α- particle fired towards a nucleus with momentum p,is r. The distance of closest approach when the momentum of α-particle is 2p is
(A). 2r
(B). 4r
(C). r/2
(D). r/4
2
1
mv
2
=
4πϵ
0 r
q
1
q
2
Initially,
2m
p
2
=
4πϵ
0
r
2e q
----(1)
Now,
2m
(2p)
2
=
4πϵ
0
r
1
2eq
----(II)
Equation (I)÷(ii)
4
1
=
r
r
1
r
1
=
4
r
Answered by
0
Answer:
The charge on an alpha particle is +2.
Hence, q=2e=3.2×10
−19
C
This alpha particle is accelerated by a potential of 2×10
6
V.
Hence, energy of the particle =qV=3.2×10
−19
×2×10
6
=6.4×10
−13
J
At closest approach, PE= total energy.
i.e.
6.4×10
−13
=
4πϵ
0
1
d
47×2e
2
Solving this, we get: d=3.384×10
−14
m
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