Answer :
the number of half-lives that would have occur to reduce a 15 g sample of Nobelium-254 down to 0.09375 g is 7.3
To calculate the number of half-lives that would have occur to reduce a 15 g sample of Nobelium-254 down to 0.09375 g, we use the formula below.
Formula:
- 2ⁿ = R/R'........... Equation 1
Where:
- n = Number of half-lives that have occured
- R = Original mass of Nobelium-254
- R' = Mass of Nobelium-254 after decay
From the question,
Given:
- R = 15 g
- R' = 0.09375 g
Substitute these values into equation 1
- 2ⁿ = 15/0.09375
- 2ⁿ = 160
- n = log160/log2
- n = 7.3 half-lives
Hence, the number of half-lives that would have occur to reduce a 15 g sample of Nobelium-254 down to 0.09375 g is 7.3
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