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Digiron [165]
3 years ago
8

Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long peri

ods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405 ✕ 1010 y. If an average lantern mantle contains 200 mg of thorium, what is its activity (in Bq)?
Physics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

The activity is 811.77 Bq

Solution:

As per the question:

Half life of Thorium, t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs

Mass of Thorium, m = 200 mg = 0.2 g

M = 232 g/mol

Now,

No. of nuclei of Thorium in 200 mg of Thorium:

N = \frac{N_{o}m}{M}

where

N_{o } = Avagadro's number

Thus

N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}

Also,

Activity is given by:

\frac{0.693}{t_{\frac{1}{2}}}\times N

A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr

A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq

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A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra
sergey [27]

Answer:

\frac{di}{dt}  = 7.31 \  A/s

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     The  number of turns is  N =  450 \  turns

      The  radius is  r =  1.17 \ cm =  0.0117 \ m

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  e  =  8.20 *10^{-6} \  V/m

Generally according to Gauss law

        \int\limits { e } \, dl  =  \mu_o *  N  *  \frac{di}{dt }  *  A

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Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                A =  \pi r ^2

=>      e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  \pi r^2

=>       \frac{di}{dt}  =  \frac{2e * x  }{\mu_o * N  * r^2}ggl;

Here  \mu_o is the permeability of free space with value

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=>     \frac{di}{dt}  =  \frac{2 *  8.20*10^{-6} *  0.0345  }{ 4\pi * 10^{-7} * 450  * (0.0117)^2}

=>      \frac{di}{dt}  = 7.31 \  A/s

6 0
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An orbiting meteor has a mass of 5,123,788 g. What is the mass of the meteor in kg?
gavmur [86]

Answer:

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Explaination:

I did 5,123,788 divided by 1000 and got 5123.788

Then I rounded to the nearest 10th place.

Hope this helps, have a great day!

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