# 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 - Physics

Numerical

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

#### Solution

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

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

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

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

Suppose after t days, the source has 10% of ""_15^32"P' nucleus and 90% of ""_15^33"P" nucleus.

Initially:

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

Number of ""_15^32"P" nucleus = 9 N

Finally:

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

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

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

"N'"/(9"N'") = (1/2)^("t"/"T"_(1//2))

"N'" = 9"N" (2)^(-1/14.3)    ....(1)

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

9"N'" = "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^(-((11"t")/(25.3 xx 14.3)))

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

(-11"t")/(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 ""_15"P"^33.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.25 | Page 464
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 25 | Page 464

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