The weight of C 27 60 O left after 10.6 years

Question

Want to see more full solutions like this?

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

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

If the room’s radius is 16.2 m, at what minimum linear speed does Quicksilver need to run to stay on the walls without sliding down? Assume the coefficient of friction between Quicksilver and the wall is 0.236.

In the comics Thor flies by spinning his hammer really fast from a leather strap at the end of the handle, letting go, then grabbing it and having it pull him. If Thor wants to reach escape velocity (velocity needed to leave Earth’s atmosphere), he will need the linear velocity of the center of mass of the hammer to be 11,200 m/s. A) If the distance from the end of the strap to the center of the hammer is 0.334 m, what angular velocity does Thor need to spin his hammer at to reach escape velocity? b) If the hammer starts from rest what angular acceleration does Thor need to reach that angular velocity in 4.16 s? c) While the hammer is spinning at its maximum speed what impossibly large tension does the leather strap, which the hammer is spinning by, exert when the hammer is at its lowest point? The hammer has a total mass of 20.0kg.

The car goes from driving straight to spinning at 10.6 rev/min in 0.257 s with a radius of 12.2 m. The angular accleration is 4.28 rad/s^2. During this flip Barbie stays firmly seated in the car’s seat. Barbie has a mass of 58.0 kg, what is her normal force at the top of the loop?

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

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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