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shtirl [24]
2 years ago
8

How long will it take a sample of lead-212 (which has a half-life of 10.64 h) to decay to one-eighth its original strength?

Physics
1 answer:
Valentin [98]2 years ago
5 0

Answer:

31.92 h

Explanation:

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 1

Amount remaining (N) = ⅛

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

⅛ = 1/2ⁿ × 1

Cross multiply

2ⁿ = 8

Express 8 in index form with 2 as the base.

2ⁿ = 2³

n = 3

Thus, 3 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Half-life (t½) = 10.64 h

Number of half-lives (n) = 3

Time (t) =?

n = t / t½

3 = t / 10.64

Cross multiply

t = 3 × 10.64

t = 31.92 h

Therefore, it will take 31.92 h for lead-212 to decay to one-eighth its original strength.

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3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
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