Share

# A Source Contains Two Phosphorous Radio Nuclides ""_15^32p(T1/2 = 14.3d) and ""_15^33p (T1/2 = 25.3d). Initially, 10% of the Decays Come from ""_15^33p. How Long One Must Wait Until 90% Do So - CBSE (Science) Class 12 - Physics

#### Question

A source contains two phosphorous radio nuclides ""_15^32P(T1/2 = 14.3d) and ""_15^33P (T1/2 = 25.3d). Initially, 10% of the decays come from ""_15^33P. How long one must wait until 90% do so?

#### Solution

Half life of ""_15^32P, T1/2 = 14.3 days

Half life of ""_15^33P, T’1/2 = 25.3 days

""_15^33P nucleus decay is 10% of the total amount of decay.

The source has initially 10% of  ""_15^33P nucleus and 90% of ""_15^33P nucleus.

Suppose after t days, the source has 10% of ""_15^32P nucleus and 90% of ""_15^33P nucleus.

Initially:

Number of ""_15^33P nucleus = N

Number of ""_15^32P nucleus = 9 N

Finally:

Number of ""_15^33P nucleus = 9N'

Number of ""_15^32P nucleus = N'

For ""_15^32P nucleus, we can write the number ratio as:

N'/(9N) = (1/2)^t/T_(1/2)

N' =9N (2)^(-1/14.3)

For ""_15^33P we can write the number ratio as:

9N' = N(2)^(-1/(25.3))  ...2

On dividing equation (1) by equation (2), we get:

1/9 = 9 xx 2^(t/25.3 - t/14.3)

1/18 = 2^(-(11t/(25.3 xx 14.3)))

log 1 - log 81 = -11t/(25.3 xx  14.3) log 2

(-11t)/(25.3 xx 14.3) = (0 - 1.908)/(0.301)

t = (25.3 xx 14.3 xx 1.908)/11 xx 0.301 ~~ 208.5 day

Hence, it will take about 208.5 days for 90% decay of ""_15P^33.

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution A Source Contains Two Phosphorous Radio Nuclides ""_15^32p(T1/2 = 14.3d) and ""_15^33p (T1/2 = 25.3d). Initially, 10% of the Decays Come from ""_15^33p. How Long One Must Wait Until 90% Do So Concept: Radioactivity - Law of Radioactive Decay.
S