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Properties of Alpha, Beta and Gamma Radiations

The metal use...

Question

The metal used for making radiation shield is:

A l

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F e

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Z n

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P b

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Solution

The correct option is C $P b$
Shielding reduces the intensity of radiation depending on thickness. A quantity known as the halving-thicknesses is used to calculate. The half value layer of lead is 1.0 cm.

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Z n (s) + C u {({N O}_{3})}_{2} (a q) \to C u (s) + Z n {({N O}_{3})}_{2} (a q)

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E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

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