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If the activity of a radioactive substance drops to 1/32 initial of its initial value in 45 years, calculate the half life period of the substance.

a.) 3 years
b.) 6 years
c.) 2 years
d.) 9 years

Respuesta :

nuralfaqih

Known :

M = Mo/32

t = 45 years

Solution :

M = Mo(1/2)^(t/T)

Mo/32 = Mo(1/2)^(45/T)

(1/2)⁵ = (1/2)^(45/T)

5 = 45/T

T = 45/5

T = 9 years

onyebuchinnaji

The half life period of the radioactive substance that dropped to 1/32 initial value is 9 years.

Half life period

The half life period of the radioactive substance is determined by using the following formula;

[tex]N = N_0(\frac{1}{2} )^{\frac{t}{t_{1/2}} }[/tex]

when, N = N₀/32, the half life period of the substance is calculated as follows;

[tex]\frac{N_0}{32} = N_0(\frac{1}{2} )^{\frac{t}{t_{1/2}} }\\\\(\frac{1}{2} )^5 = (\frac{1}{2} )^{\frac{t}{t_{1/2}} }\\\\5 = \frac{t}{t_{1/2}}\\\\5 = \frac{45}{t_{1/2}} \\\\t_{1/2} = \frac{45}{5} \\\\t_{1/2} = 9 \ years[/tex]

Thus, the half life period of the radioactive substance that dropped to 1/32 initial value is 9 years.

Learn more about half life here: https://brainly.com/question/2320811

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