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The isotope 208 Tl undergoes $\beta$ decay with a half-life of 3.1 $\mathrm{min}$ . (a) What isotope is produced by the decay?(b) How long will it take for 99.0$\%$ of a sample of pure $^{208} \mathrm{T}$ to decay? (c) What percentage of a sample of pure $^{208}$ Tl remains un-decayed after 1.0 $\mathrm{h} ?$

a) $_{82}^{208} \mathrm{Pb}$b) 20.9 $\mathrm{min}$c) $1.8 \times 10^{-40 / 0}$

Chemistry 102

Chapter 21

Nuclear Chemistry

University of Central Florida

University of Kentucky

Lectures

00:29

In nuclear physics, radioa…

08:39

In physics, there are thre…

03:03

43.20. The isotope 90 Sr u…

0:00

In a comparison of two rad…

06:51

A sample of radioactive $_…

A certain radioactive nucl…

04:01

A sample of tritium-3 deca…

04:17

The half-life of $^{131}$ …

03:09

$^{209}_{79}$ $\mathrm{P}o…

04:18

The half-life of 131 $\mat…

03:16

A sample of tritium- 3 dec…

01:02

A nuclide has a half-life …

04:05

(II) A sample of $^{233}_{…

05:27

The radioactive isotope 19…

01:43

What is the half-life of i…

06:02

The radioactive isotope $^…

02:55

SSM WWW A radioactive isot…

02:15

How many half-lives must e…

01:51

What nuclide is produced i…

05:25

The radioactive isotope th…

05:02

14:10

Radium-226, which undergoe…

the half life is defined as the time required for the activity off radioactive element toe fall to its half offense in the selectivity that is known as half life delay show off radioactive substance left is calculated by the given formula. Let's market as number one Bertie is time. Lambda is decay constant and end by and not is the ratio off initial concentration and not and become situation at time T for the party. Now the mass off number off beat a particle is zero and atomic number is minus one. So the daughter nucleus will have atomic number off 82 on the mass 208 for bite be the half. Life is 3.1 minute. We can substitute this value into the equation, substituting the values we get. Lando, that is, we get glammed up. That is the gay constant as going toe per minute. Now, the fish off initial amount on DDE on dhe amount at Time T can be calculated putting in the values we have ratio equal to 0.1 Now we substitute all off the given values in this equation. It gives us time that comes out to between people in nine minutes for part C, The percentage left is calculated as follows. We're time a 60 minute, substituting the values, substituting the racial values we have re showdown out, Toby. 1.8 into 10 the super minus six. The percentage off radioactive material left can be calculated. That is 1.8 into 10 days, super minus 4%.

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