Question

Considering that the alpha-particles carry average kinetic energy of 2.00 x 1010 J,
calculate the maximum size of the gold nucleus. [Atomic number of gold is 79 and
e= 1.60 x 10-19C] [3]
Explain why the radius of the gold nucleus must be much smaller than the value
calculated in above.

Answer 1

Explanation:

number of basic units of charge, e. ... each droplet are all integer multiples of -1.60χ10-19 C. We therefore state ... Atomic and Nuclear Physics – Page 2 of 14 ... Answer is 8.5 × 1010 J.

Answer 2

Answer: 8.5 × 1010 J.

The radius of the gold nucleus must be much smaller than the value

Explanation:

A number of basic units of charge, e. ... each droplet are all integer multiples of $-1.60*10-19$ C.

We, therefore, state that the atomic number of gold is 79 with e= 1.60 x 10-19C

Due, to this the radius of gold is smaller as 2*1010J<8,5 x 1010J

The answer is 8.5 × 1010 J.

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