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larisa [96]
2 years ago
8

The isotope 198 au (atomic mass = 197.968 u) of gold has 79 a half-life of 2.69 days and is used in cancer therapy. what mass (i

n grams) of this isotope is required to produce an activity of 315 ci?
Physics
1 answer:
Kaylis [27]2 years ago
6 0

391.12 * 10^{16} mass is required for activity of 315 ci

The activity of the radioactive substance is the number of disintegrations taking place in a radioactive sample.

Isotopes are the members of an element having the same number of protons but the different number of neutrons.

the activity of the radioactive sample is given as:

A = λ * N

where λ is the decay constant and N is the number of disintegrations.

activity is given as 315 ci

A = 315 * 3.7 * 10^{10}

   = 1165.5 *  10^{10} Bq

λ = 0.693 / t_{1/2}

   = 0.693 / 2.69 * 24 * 3600

   = 2.98 * 10^{-6} t^-1

Therefore, N = A /  λ

                      = 1165.5 *  10^{10} /  2.98 * 10^{-6}

                  N = 391.12 * 10^{16}

391.12 * 10^{16} mass is required for activity of 315 ci

For more information click on the link below:

brainly.com/question/25746629

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In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc
Harman [31]

Answer:

A=\frac{A_0}{\sqrt 3}

Explanation:

Initial amplitude=A_0

We are given that

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We have to find the amplitude after 1 s.

We know that amplitude at any time t

A=A_0e^{-\alpha t}

Using the formula

\frac{A_0}{3}=A_0e^{-2\alpha}

\frac{1}{3}=e^{-2\alpha}

3=e^{2\alpha}

ln 3=2\alpha

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e^{\alpha}=\sqrt 3}

When t=1 s

A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}

8 0
4 years ago
a 10kg ball is thrown into the air. it is going 3m/s when thrown. How much potential energy will it have at the top?
Alexandra [31]

Answer:

45 J

Explanation:

Assuming the level at which the ball is thrown upwards is the ground level,

We can use the equations of motion to obtain the maximum height covered by the ball and then calculate the potential energy

u = initial velocity of the ball = 3 m/s

h = y = vertical distance covered by the ball = ?

v = final velocity of the ball at the maximum height = 0 m/s

g = acceleration due to gravity = -9.8 m/s²

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19.6y = 9

y = (9/19.6)

y = 0.459 m

The potential energy the ball will have at the top of its motion = mgh

mgh = (10)(9.8)(0.459) = 45 J

Hope this Helps!!!

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Please help soon!!!!
anyanavicka [17]
The Answer: Parallel
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