1.

The half-life period of Pb^(210) is 22 years. If 2 g of Pb^(210) is taken, then after 11 years how much of Pb^(210) will be left

Answer»

1.414 g
2.428 g
3.442 g
4/456 g

Solution :`N_(t) = N_(0) ((1)/(2))^(2) [ :' t_(1//2) = 22 " YEARS, " T = 11 " years" , N_(0) = 2, N_(t) = ?]`
`T = t_(1//2) xx N, 11 = 22 xx n or n = (11)/(22) = (1)/(2)`
`:. N_(t) = 2g xx ((1)/(2))^(1//2) = 1.414g`


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