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Rufina [12.5K]
3 years ago
6

In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in

s) of the material
Physics
1 answer:
almond37 [142]3 years ago
7 0

Answer:

<h2>2113 seconds</h2>

Explanation:

The general decay equation is given as N = N_0e^{-\lambda t} \\\\, then;

\dfrac{N}{N_0} = e^{-\lambda t} \\ where;

N/N_0 is the fraction of the radioactive substance present = 1/16

\lambda is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

\frac{1}{16} = e^{-\lambda(8450)}  \\\\Taking\ ln\ of \both \  sides\\\\ln(\frac{1}{16} ) =  ln(e^{-\lambda(8450)})  \\\\\\ln (\frac{1}{16} )  = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328

Half life f the material is expressed as t_{1/2} = \frac{0.693}{\lambda}

t_{1/2} = \frac{0.693}{0.000328}

t_{1/2} = 2,112.8 secs

Hence, the half life of the material is approximately 2113 seconds

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Answer:

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1hp is written in electric motor what does it mean​
Angelina_Jolie [31]

Explanation:

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A styrofoam container used as a picnic cooler contains a block of ice at 0°C. If 325 g of ice melts in 1 hour, how much heat ene
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Answer:

30.0625 W

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3 years ago
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts
prohojiy [21]

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

N=N_0e^{-\lambda t }

.70446 =e^{-\lambda t }

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

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7 0
3 years ago
Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the d
uysha [10]

Answer:

Y = 3.87 x 10⁻³ m = 3.87 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

Y = \frac{\lambda L}{d}

where,

Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

d = slit width = 0.3 mm = 3 x 10⁻⁴ m

Therefore,

Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}

<u>Y = 3.87 x 10⁻³ m = 3.87 mm</u>

3 0
3 years ago
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