Question

^​38₁₆S decays by beta emission to ³⁸₁₇​Cl. This reaction has a half‐life of 2.87 hours. The ³⁸₁₇​Cl subsequently decays by beta emission to ³⁸₁₈​Ar. This reaction has a half‐life of 37.3 min.

a) How much will remain of an initially pure sample of 3.5 mg ³⁸₁₆S after 9.35 hours?
b) How many mg of ³⁸₁₇​Cl will be present after 9.35 hours?
c) How many mg of ³⁸₁₈​Ar will be present after 9.35 hours?

Answer 1

a) ₃₈¹⁶S ----> β− +₃₈¹⁷ Cl

Calculate k for reaction

Half‐life = 2.87 hours

K = 0.693/2.87 hours = 0.2415/h

Pure sample C_o = 3.5 mg

t = 9.35 hours

ln ( C / C_o ) = - k t

ln ( C / Co ) = - 0.2415/h x 9.35 hours = -2.258

(C/ C_o) = e^(-2.258) = 0.105

(C/3.5 mg) = 0.105

C = (3.5g x 0.105) = 0.3675 mg

After 9.35 hours 0.3675 mg ₃₈¹⁶S will remain.

b)

         3.1325 g ₃₈¹⁷ Cl will subsequently decays by beta emission to ³⁸₁₈​Ar

³⁸₁₇​Cl ----> β− + ³⁸₁₈​Ar

Half‐life = 37.3 min

K = 0.693/37.3 min = 0.01858/min

Pure sample Co ³⁸₁₇​Cl = 3.1325 mg

ln ( C / C_o ) = - k t

ln ( C / Co ) = - 0.01858/min x (9.35 x 60) min = -10.423

(C/ C_o) = e^(-10.423) = 2.97 x 10^-5

C/3.1325 mg = 2.97 x 10^-5

C = (3.1325 mg x 2.97 x 10^-5 )

= 9.30 x 10^-5 mg ₃₈¹⁷ Cl will be present after 9.35 hours

c)

³⁸₁₈​Ar = 3.1325 mg - 9.30 x 10^-5

3.132 mg of ³⁸₁₈​Ar will be present after 9.35 hours