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6 Radioactivity

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Question 145
F

The equation represents alpha decay.

84210Po→ZAPb+24He {}^{210}_{84}\mathrm{Po}\rightarrow{}^{A}_{Z}\mathrm{Pb}+{}^{4}_{2}\mathrm{He} 84210​Po→ZA​Pb+24​He

Which values complete the equation?

[1]
A

A=208A=208A=208 and Z=82Z=82Z=82

B

A=206A=206A=206 and Z=82Z=82Z=82

C

A=210A=210A=210 and Z=82Z=82Z=82

D

A=206A=206A=206 and Z=84Z=84Z=84

Markscheme

6 Radioactivity Questions

  1. GCSE
  2. /Physics
  3. /6 Radioactivity

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