A sample of a radioactive substance decayed to 91.5% of its original amount after a year. (Round your answers to two decimal places.) (a) What is the half-life of the substance

Answer: t½ = 7.80

Explanation:

This question is lacking the (b) part and source similar online are not the same since some values has been altered. However I'll answer the one present here.

Nt = No. e^-yt

Note : y is used to represent the greek alphabet lambda

Nt = 91.5

No = 100

t = 1 year

Nt = No. e^-yt

91.5 = 100e^-yt

e^-y = 91.5/100

Take ln on both sides

In (e^y) = ln (100/91.5)

y = ln (1.0929)

y = 0.0888

t½ = 0.693 / y

t½ = 0.693 / 0.0888

t½ = 7.80