The half life of 231Pa is 3.25 x 10^4 years. How much of an initial 10.40 microgram sample remains after 3.25 x 10^5 years?
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Answer:
Given t1/2 for 231Pa = 3.25 × 104 yr
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1λ= 2.1323 × 10-5 yr-1
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1λ= 2.1323 × 10-5 yr-1Now radioactive disintegration follows first order rate law so we get a equation
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1λ= 2.1323 × 10-5 yr-1Now radioactive disintegration follows first order rate law so we get a equationNt = N0e−λt
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1λ= 2.1323 × 10-5 yr-1Now radioactive disintegration follows first order rate law so we get a equationNt = N0e−λtWhere N0 = Amount initially present = 10.40 microgram
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1λ= 2.1323 × 10-5 yr-1Now radioactive disintegration follows first order rate law so we get a equationNt = N0e−λtWhere N0 = Amount initially present = 10.40 microgramNt = amount present after time t = need to be calculate
Given t1/2 for 231Pa = 3.25 × 104 yrThe radioactive disintegration constant(λ) is related by the equation as,λ= 0.693/t1/2λ= 0.693/(3.25 × 104) yr-1λ= 2.1323 × 10-5 yr-1Now radioactive disintegration follows first order rate law so we get a equationNt = N0e−λtWhere N0 = Amount initially present = 10.40 microgramNt = amount present after time t = need to be calculateλ= radioactive disintegration constant = λ= 2.1323 × 10-5 yr-1
t= time = 3.25 × 105 yr
t= time = 3.25 × 105 yrNow from the above euation we get
t= time = 3.25 × 105 yrNow from the above euation we getNt = N0e−λt
t= time = 3.25 × 105 yrNow from the above euation we getNt = N0e−λtNt = 10.40e−(2.1323 × 10-5 × 3.25 × 105) microgram
t= time = 3.25 × 105 yrNow from the above euation we getNt = N0e−λtNt = 10.40e−(2.1323 × 10-5 × 3.25 × 105) microgramNt = 10.40× e-6.9299 microgram
t= time = 3.25 × 105 yrNow from the above euation we getNt = N0e−λtNt = 10.40e−(2.1323 × 10-5 × 3.25 × 105) microgramNt = 10.40× e-6.9299 microgramNt = 10.40 × 9.78 × 10-4 microgram
t= time = 3.25 × 105 yrNow from the above euation we getNt = N0e−λtNt = 10.40e−(2.1323 × 10-5 × 3.25 × 105) microgramNt = 10.40× e-6.9299 microgramNt = 10.40 × 9.78 × 10-4 microgramNt = 0.0102 microgram
REQUIRED ANSWER : 0.0102mg
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