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IIT JEE Advanced 2012 Paper 1
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© examsnet.com
Question : 18
Total: 60
A proton is fired from very far away towards a nucleus with charge Q = 120e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is _____.
(Take the proton mass,
m
p
= (5/3) ×
10
−
27
kg; h/e = 4.2 ×
10
−
15
J.s/C;
1
4
π
ε
0
= 9 ×
10
9
m/F; 1 fm =
10
15
m.)
Your Answer:
Validate
Solution:
We have
0
+
1
2
m
v
2
=
K
(
Q
)
e
10
×
10
−
15
=
K
(
120
e
)
e
10
×
10
−
15
1
2
×
5
3
×
10
−
27
v
2
=
9
×
10
9
×
120
×
(
1.6
×
10
−
19
)
2
10
×
10
−
15
v =
9
×
6
×
10
9
×
120
×
2.56
×
10
−
38
50
×
10
−
42
=
√
331.776
×
10
13
The de Broglie wavelength (in units of fm) of the proton at its start is calculated as follows:
λ =
h
m
v
=
4.2
×
10
−
15
×
1.6
×
10
−
19
5
3
×
10
−
27
×
√
331.776
×
10
13
=
4.2
×
4.8
×
10
−
34
57.6
×
5
×
10
−
21
= 0.07 ×
10
−
13
=
7
×
10
−
15
= 7 fm
Hence, the correct answer is 7.
© examsnet.com
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