HI, I need help my my edg. Can you help me? I will give crown.<br> :D

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

2 years ago

ASAP something that flows into a substance

Radda [10]

Answer:

can oxygen exist as a liquid and solid

7 0

2 years ago

Please help! need the answer fast. Determine the amplitudes and the wave-lengths of each of the three waves.

Diano4ka-milaya [45]

A amplitude = 0.50 m and wave-length = 1.0 m

B amplitude = 0.40 m and wave-length = 2.0 m

C amplitude = 0.60 m and wave-length = 2.0 m

3 0

2 years ago

What would be the best flow rate of o2 for a patient with copd?

Step2247 [10]

When we say COPD, this means Chronic Obstructive Pulmonary Disorder. COPD includes a group of lung diseases which have progressed into a worse condition. One example of this is emphysema. In cases of COPD, they should be provided oxygen therapy and the best flow rate for patients with this disorder is <span>8-10 L/min, high-flow, via nasal prong. Hope this helps.</span>

4 0

2 years ago

What is the empirical formula of a vanadium (i) oxide given that 20.38 grams of vanadium combines with oxygen to form 23.58 gram

oksano4ka [1.4K]

V₂O is the empirical formula of a vanadium (i) oxide given that 20.38 grams of vanadium combines with oxygen to form 23.58 grams of the oxide.

The simplest whole number ratio of the atoms of the elements in a specific compound is shown by the empirical formula for that compound. When the mass of each component of a compound or its % composition by mass is known, an empirical formula may typically be computed.

% composition is equal to (mass of an element minus mass of the compound) / 100%.

Given

Vanadium weighs 20.38 g.

The oxide's mass is 23.58 g.

% Vanadium composition: (20.38 g/23.58 g) 100%

= 86.43%

Oxygen mass equals 23.58 g minus 20.38 g

= 3.2 g

100% of oxygen's composition is (3.2/g/23.58 g)

= 13.57%

Atomic mass and number are related by composition percentage.

Vanadium's atomic weight is 50.94 g/mol.

Atomic weight of V = 86.43 / 50.94

= 1.6967

Oxygen has an atomic mass of 16 g/mol.

O atom count is 13.57/16.

= 0.8481

Now on calculation the ratio of Vanadium to the oxygen is

1.6967/0.8481 = 2/1

Ratio of whole numbers is 2:1

Thus V₂O is the empirical formula .

Learn more about empirical formula here brainly.com/question/13058832

#SPJ4

6 0

6 months ago

HI, I need help my my edg. Can you help me? I will give crown. :D