The activity of a radioactive material is 2.56 × 10-3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 × 10-5 Ci?

A0 = 2.56 × 10-3Ci A = 2 × 10-15Ci A = <!- [if gte mso 9]><xml> </xml><! [endif]-> A 0 2 n <!- [if gte mso 9]><xml> </xml><! [endif]-> ⇒ 2 × 1 0 − 5 = 2 . 5 6 × 1 0 − 3 2 n 2n = 128 n = 7 t = 7 × t1/2 = 7 × 5 = 35 days

A0 = 2.56 * 10-3Ci A = 2 * 10-15Ci A = <!- [if gte mso 9]><xml> </xml><! [endif]-> A 0 2 n <!- [if gte mso 9]><xml> </xml><! [endif]-> ? 2 * 1 0 ? 5 = 2 . 5 6 * 1 0 ? 3 2 n 2n = 128 n = 7 t = 7 * t1/2 = 7 * 5 = 35 days

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Physics NCERT Exemplar Solutions Class 11th Chapter Five Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Physics Nuclei

The activity of a radioactive material is 2.56 × 10^-³ Ci. If the half life of the material is 5 days, after how many days the activity will become 2 × 10^-⁵ Ci?

Option 1 -

30 days

Option 2 -

35 days

Option 3 -

40 days

Option 4 -

25 days

3 Views | Posted 2 months ago

Asked by Shiksha User

2 Answers

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

A₀ = 2.56 × 10^-3C_i

A = 2 × 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 \times 1 0^{- 5} = \frac{2.56 \times 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

&nbs

...more

A₀ = 2.56 × 10^-3C_i

A = 2 × 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 \times 1 0^{- 5} = \frac{2.56 \times 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 × t_1/2

= 7 × 5 = 35 days

less

A0 = 2.56 × 10-3Ci              A = 2 × 10-15Ci              A = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814257605"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mi>n</mi> </mrow> </msup> </mrow> </mfrac> </mrow> </math>               <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814257606"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mo>⇒</mo> <mn>2</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>5</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>5</mn> <mn>6</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mi>n</mi> </mrow> </msup> </mrow> </mfrac> </mrow> </math>                       2n = 128                      n = 7                      t = 7 × t1/2                      = 7 × 5 = 35 days

0 Upvotes 0 Downvotes

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$? 2 * 1 0^{? 5} = \frac{2.56 * 1 0^{? 3}}{2^{n}}$

2ⁿ = 128

n = 7

&nbs

...more

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$? 2 * 1 0^{? 5} = \frac{2.56 * 1 0^{? 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 * t_1/2

= 7 * 5 = 35 days

less

A0 = 2.56 * 10-3Ci              A = 2 * 10-15Ci              A = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814257605"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mi>n</mi> </mrow> </msup> </mrow> </mfrac> </mrow> </math>               <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814257606"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mo>? </mo> <mn>2</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>5</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>5</mn> <mn>6</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mi>n</mi> </mrow> </msup> </mrow> </mfrac> </mrow> </math>                       2n = 128                      n = 7                      t = 7 * t1/2                      = 7 * 5 = 35 days