The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) is

75.8 fm

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7.58 fm

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758 fm

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0.758 fm

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Solution

The correct option is B
7.58 fm

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Similar questions

Q. The distance of the closest approach when a

5.0 MeV

proton approaches a gold nucleus is

(Z = 79)

Q. The maximum K.E found in alpha particle of natural origin is

7.7

MeV. Find the closest distance of approach for the gold nucleus

(Z = 79

for gold

)

[Note : At closest distance of approach, kinetic energy is converted into potential energy]

Q. A

1 M e V

proton is sent against a gold leaf

(Z = 79)

. Calculate the distance of closest approach for head-on collision.

Q. A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.

An alpha particle of $5 M e V$ at a large distance proceeds towards a gold nucleus $(Z = 79)$ to make a head on collision. The closest distance of approach from the centre of gold nucleus is:

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The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) is

The correct option is B 7.58 fm

The correct option is B
7.58 fm