The weight of C 27 60 O left after 10.6 years

Question

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Answer 1

Question

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

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Answer 2

Question

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

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Answer 3

Question

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

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Answer 4

Question

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

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Answer 5

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

Answer 6

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

Answer 7

Chapter 30, Problem 1OQ

To determine

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Answer 8

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

b)20µg

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Half -life of ${}_{27}^{60}C O$ , t_1/2 = 5.3 years

Initial weight of ${}_{27}^{60}C O$ , N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

$t_{1 / 2} = \frac{0.693}{λ}$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$N = N_{0} e^{- λ t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{27}^{60}C O$ can be found as,

$\begin{array}{l} λ = \frac{0.693}{t_{1 / 2}} \\ = \frac{0 .693}{5.3} \\ = 0 {.1307 year}^{- 1} \end{array}$

Now from eqn.2, amount of ${}_{27}^{60}C O$ left after 10.6 years can be calculated

$\begin{array}{l} N = 80 \times e^{- 0.1307 \times 10.6} \\ = 20 .017 μ g \end{array}$

Answer 12

Ch. 30 - Prob. 1OQ Ch. 30 - Prob. 1Q Ch. 30 - Prob. 2Q Ch. 30 - Prob. 3Q Ch. 30 - Prob. 4Q Ch. 30 - Prob. 5Q Ch. 30 - Prob. 6Q Ch. 30 - Prob. 7Q Ch. 30 - Prob. 8Q Ch. 30 - Prob. 9Q

The weight of C 27 60 O left after 10.6 years

Concept explainers

The weight of ${}_{27}^{60}C O$ left after 10.6 years

Answer to Problem 1OQ

Explanation of Solution

Want to see more full solutions like this?

Chapter 30 Solutions

The weight of C 27 60 O left after 10.6 years

Concept explainers