The half-life of K-42 is 12.360 hours.
62 hours is just about 5 half-lives, so we'd expect about 1/32 of the original amount to remain after that time. (32 = 2^5)
In fact, it's 12.36 / 62 = 5.016 half-lives.
2^5.016 = 32.36
So if you started with 32.36g in the original sample, it would have decayed down to 1g after 5.016 half-lives.
According to the question,
As we know,
→ [tex]n = \frac{t}{K_{1/2}}[/tex]
[tex]= \frac{62}{12.36}[/tex]
[tex]= 5.016[/tex]
hence,
The original sample will be:
→ [tex]N = N_o(\frac{1}{2} )^n[/tex]
or,
→ [tex]N_o = N\times 2^n[/tex]
By substituting the values, we get
[tex]= 1\times (2)^{5.016}[/tex]
[tex]= 32 \ g[/tex]
Thus the above approach is right.
Learn more K half life here:
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