Does anyone know how to find the answer key to this?

•A radioactive material A (decay constant λA) decays into a material B (decay constant λB) and then into material C (decay const

KiRa [710]

Answer:

The amount of C remaining after time t is

$N_C__{R}} =N_D = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C } ]$

Explanation:

We can represent the decay sequence as

$A \to B \to C \to D$

The reason we added D is because we are told from the question that C is also radioactive so it has the tendency to decay

Generally for every decay the remaining radioactive element can be obtained as

$N = N_0 -N_0 e^{- \lambda t}$

Where N is the amount of the remaining radioactive material

$N_0$ is the original amount amount of the radioactive material before decay

and $\lambda$ is the decay constant

Now for the decay from $A \to B$ amount of radioactive element B formed from A after time t can be obtained as

$N_b = N_0 -N_0 e^{- \lambda_A t}$

Where $\lambda _A$ is the decay constant of A

Now for the decay from $B \to C$ amount of radioactive element C formed from A after time t can be obtained as

$N_c = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{\lambda_A t})e^{-\lambda_B t}$

Where $\lambda _B$ is the decay constant of B

Now for the decay from $C \to D$ amount of radioactive element D formed from A after time t can be obtained as

$N_C__{R}} =N_D = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C } ]$

So this amount of D is the reaming amount of the radioactive material C

6 0

2 years ago

Please guys help me plz for god sake plz plz guys plz plz plz i will mark u as a brainlist plz

victus00 [196]

Answer:

Explanation:

The complete sequence of motions and activities required to complete one work cycle.

8 0

2 years ago

Light rays in a material with index of refraction 1.31 can undergo total internal reflection when they strike the interface with

3241004551 [841]

Answer:

The refractive index of the material is 1.28.

Explanation:

It is given that,

Refractive index of medium 1, n₁ = 1.31

Critical angle, $\theta_c=78.3$

At critical angle rays will reflects at 90 degrees. Using Snell's law as:

$n_1sin\theta_1=n_2sin\theta_2$

At critical angle, $n_1sin\ \theta_c=n_2\ sin90$

$n_1sin\ \theta_c=n_2$

$n_2=1.31\times sin(78.3)$

$n_2=1.28$

So, the refractive index of the material is 1.28. Hence, this is the required solution.

3 0

2 years ago

A force of 8 N accelerates by 4 m/s^2. What would be the amount of force needed to give a final acceleration of 5.3 m/s^2

Elis [28]

Answer:

10.6 N

Explanation:

F = ma

8 = m * 4

m = 2 kg

F = 2 * 5.3

F = 10.6 N

4 0

2 years ago

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

aleksandr82 [10.1K]

Answer:

Explanation:

We shall apply Doppler's effect to solve the problem .

Formula for apparent frequency for a source of sound approaching an observer is as follows .

f₁ = f₀ V / (V - v )

where f₁ and f₀ are apparent and real frequency of source , V and v is velocity of sound and velocity of approaching source respectively .

Putting the given values and knowing that speed of sound is 340 m /s

f₁ =346x 340 / (340 - 39.6 )

f₁ = 391.6 Hz

In case of receding train , the formula is

f₂ = f₀ V / (V + v )

Putting the values

f₂ = 346x 340 / (340 + 39.6 )

= 309.9 Hz

Change in frequency = 391.6 - 309.9

= 81.7 Hz .

4 0

2 years ago