Compute the activity of one gram of Th 232 whose decay constant is 1 .58 × 10 − 18 s − 1 . Take Avogadro’s number 6 .02 × 10 23 mole . ( 1 μ c = 3 .7 × 10 4 disintegrations s )

Number of atoms in 1 gram of Thorium = 6 . 023 × 10 23 232 = 2 . 59 × 10 21 atoms Activity R = − d N d t = λ N ⇒ R = 1 . 58 × 10 - 18 × 2 . 59 × 10 21 = 4092 . 2 disintegrations per second 1 μ c = 3 .7 × 10 4 disintegrations s 4092 . 2 disintegrations per second = 4092 . 2 3 . 7 × 10 4 = 0 . 11 Hence, activity is 0 .11 μc (microcurie).

Compute the activity of one gram of Th 232 whose decay constant is 1 .58 × 10 − 18 s − 1 . Take Avogadro’s number 6 .02 × 10 23 mole . ( 1 μ c = 3 .7 × 10 4 disintegrations s )

Number of atoms in 1 gram of Thorium = 6 . 023 × 10 23 232 = 2 . 59 × 10 21 atoms Activity R = − d N d t = λ N ⇒ R = 1 . 58 × 10 - 18 × 2 . 59 × 10 21 = 4092 . 2 disintegrations per second 1 μ c = 3 .7 × 10 4 disintegrations s 4092 . 2 disintegrations per second = 4092 . 2 3 . 7 × 10 4 = 0 . 11 Hence, activity is 0 .11 μc (microcurie).

2 .0 × 10 − 5 gram of Au 200 has a decay rate (activity) of 60 curie. Compute its half-life. Take Avogadro’s number 6 .02 × 10 23 / mole , 1 curie = 3 . 7 × 10 10 disintegrations per second, log e 2 = 0 . 693

We know, - d N d t = λ N ⇒ - d N N d t = λ . . . . . . . . . . . . . . . . . . 1 Given, - d N d t = 60 curie ⇒ - d N d t = 60 × 3 . 7 × 10 10 disintegrations per second Number of atoms in 1 g of gold = 6 . 02 × 10 23 200 Number of atoms in 2 . 0 × 10 - 5 g of gold, N = 6 . 02 × 10 23 200 × 2 . 0 × 10 - 5 Decay constant in terms of half life ( T ) is given by, λ = log e 2 T . . . . . . . . . . . . . . . . . . . 2 From equation 1 and 2 - d N N d t = log e 2 T ⇒ T = log e 2 × N ( - d N / d t ) ⇒ T = 0 . 693 × 6 . 023 × 10 23 × 2 × 10 - 5 200 × 60 × 3 . 7 × 10 10 = 1 . 88 × 10 4 Hence, half life is 1 .88 × 10 4 seconds.

1 gram Cs 137 decays with a half-life of 30 years. If the initial activity of Cs 137 is 1 .0 mc (millicurie), find number of atoms left after 5 years. Take ( 0 .5 ) 1 / 6 = 0 .89 . Take Avogadro's number = 6 . 023 × 10 23

Number of atoms in 1 g of Cs 137 = 6 . 023 × 10 23 137 Number of half lives = 5 30 = 1 6 By Rutherford and Soody law, N N 0 = 1 2 n N = 6 . 023 × 10 23 137 × ( 0 . 5 ) 1 6 N = 3 .91 × 10 21 Hence, number of atoms left after 5 years is 3 .91 × 10 21 .

1 gram Cs 137 decays with a half-life of 30 years. If the initial activity of Cs 137 is 1 .0 mc (millicurie), find the activity after 5 years. The Avogadro’s number is 6 .02 × 10 23 per mole. Take ( 0 .5 ) 1 / 6 = 0 .89 .

We know, R R 0 = N N 0 = 1 2 n Where n is number of half lives in given time. n = 5 30 = 1 6 R R 0 = 1 2 1 6 ⇒ R = 1 × ( 0 . 5 ) 1 6 = 1 × 0 . 89 Hence, activity after 5 years is 0 . 89 mc .

EASY

12th ICSE

IMPORTANT

Earn 100

Compute the activity of one gram of ${Th}^{232}$ whose decay constant is $1 .58 \times 10^{- 18} s^{- 1}$ . Take Avogadro’s number $6 .02 \times \frac{10^{23}}{mole}$ . ( $1 μ c = 3 .7 \times 10^{4} \frac{disintegrations}{s}$ )

Important Questions on Radioactivity

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12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>QUESTIONS>Q 22

The half-life of a radioactive material is $40$ days. What time will it take for the activity to drop to $30 %$ of its initial value? $\log_{e} 2 = 0 .6931$ and $\log_{e} (\frac{10}{3}) = 1 .20$ .

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>QUESTIONS>Q 23

2 .0 \times 10^{- 5}

gram of

{Au}^{200}

has a decay rate (activity) of

60

curie. Compute its half-life. Take Avogadro’s number

6 .02 \times 10^{23} / mole

1

curie

= 3.7 \times 10^{10}

disintegrations per second,

\log_{e} 2 = 0.693

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>QUESTIONS>Q 24

1

gram

{Cs}^{137}

decays with a half-life of

30

years. If the initial activity of

{Cs}^{137}

1 .0 mc

(millicurie), find number of atoms left after

5

years. Take

(0 .5)^{1 / 6} = 0 .89

. Take Avogadro's number

= 6.023 \times 10^{23}

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>QUESTIONS>Q 24

1

gram

{Cs}^{137}

decays with a half-life of

30

years. If the initial activity of

{Cs}^{137}

1 .0 mc

(millicurie), find the activity after

5

years. The Avogadro’s number is

6 .02 \times 10^{23}

per mole. Take

(0 .5)^{1 / 6} = 0 .89

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>QUESTIONS>Q 25

Calculate the mass of $U^{234}$ in a material having an activity of one curie. Take Avogadro’s number $6 .02 \times 10^{23} / mole$ . ( $1 c = 3.7 \times 10^{10}$ disintegrations per second)

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>QUESTIONS>Q 26

Calculate the mass of

{Co}^{60}

required to provide a radioactive source of

8.0

curie strength. The half-life of

{Co}^{60}

5.3

years. Take Avogadro’s number

6 .02 \times 10^{23} {mole}^{- 1}

. (

\log_{e} 2 = 0.693

)

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>For DIFFERENT COMPETITIVE EXAMINATIONS>Q 1

A $280$ days old radioactive sample shows an activity of $6000$ disintegrations per second (dps). The activity decreases to $3000 dps$ in the next $140$ days. The original activity of the sample was:

EASY

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 15 - Radioactivity>For DIFFERENT COMPETITIVE EXAMINATIONS>Q 2

A radioactive sample consists of two distinct species having equal number of atoms initially. The mean-life time of one species is $t$ and that of the other is $5 t$ . The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represents the form of this plot?

Compute the activity of one gram of Th232 whose decay constant is 1.58×10−18s−1. Take Avogadro’s number 6.02×1023mole. (1μc=3.7×104 disintegrationss)

Important Questions on Radioactivity

Learn and Prepare for any exam you want !

Compute the activity of one gram of ${Th}^{232}$ whose decay constant is $1 .58 \times 10^{- 18} s^{- 1}$ . Take Avogadro’s number $6 .02 \times \frac{10^{23}}{mole}$ . ( $1 μ c = 3 .7 \times 10^{4} \frac{disintegrations}{s}$ )