Question

19. The half-life of 231Pa is 3.25 x 104 years. How much of an initial 10.40 microgram sample remains after 3.25 x 10 years? (a) 0.0120 micrograms (b) 0.240 micrograms (c) 2.18 micrograms (d) 0.0102 micrograms (e) 1.04 micrograms

Answer 1

Given t_1/2 for ²³¹Pa = 3.25 × 10⁴ yr

The radioactive disintegration constant(λ) is related by the equation as,

λ= 0.693/t_1/2

λ= 0.693/(3.25 × 10⁴) yr^-1

λ= 2.1323 × 10^-5 yr^-1

Now radioactive disintegration follows first order rate law so we get a equation

N_t = N₀e^−λt

Where N₀ = Amount initially present = 10.40 microgram

N_t = amount present after time t = need to be calculate

λ= radioactive disintegration constant = λ= 2.1323 × 10^-5 yr^-1

t= time = 3.25 × 10⁵ yr

Now from the above euation we get

N_t = N₀e^−λt

N_t = 10.40e−(2.1323 × 10^-5 × 3.25 × 10⁵) microgram

N_t = 10.40× e^-6.9299 microgram

N_t = 10.40 × 9.78 × 10^-4 microgram

N_t = 0.0102  microgram

Hence (d) is the answer.

Answer 2

0.0102 micrograms

Answer 3

Answer 4