Please help me to solve this questions Homework (2)124The half-life of the isotope (¹3Cs) is (30.8 sec), Determine:551. Number of nuclei in (6.2 µg) of this isotope?2. Number of nuclei will be present after (1.2 min)?3. Activity at this time? Homework (3)Consider the (¹C) nucleus:1. Write down its nucleon's configuration according to the shell model.2. Find its nuclear spin-parity.

Given:Half-life of ,So, the mass of Cs isotope= 124 amu=(1) The number of nuclei in 6.2 micrograms…

Answered: 124 The half-life of the isotope (¹3Cs)…

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

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Homework (2)
124
The half-life of the isotope (¹3Cs) is (30.8 sec), Determine:
55
1. Number of nuclei in (6.2 µg) of this isotope?
2. Number of nuclei will be present after (1.2 min)?
3. Activity at this time?

Homework (3)
Consider the (¹C) nucleus:
1. Write down its nucleon's configuration according to the shell model.
2. Find its nuclear spin-parity.

Expert Solution

Step 1: Question Homework 2:

Given:

Half-life of $C presubscript 55 presuperscript 124 s$ , $t subscript bevelled 1 half end subscript equals 30.8 s e c$

So, the mass of Cs isotope= 124 amu= $124 cross times 1.66 cross times 10 to the power of negative 24 end exponent g$

(1) The number of nuclei in 6.2 micrograms is $N subscript 0 equals fraction numerator 6.2 cross times 10 to the power of negative 6 end exponent over denominator 124 cross times 1.66 cross times 10 to the power of negative 24 end exponent end fraction equals 30.120 cross times 10 to the power of 15$

(2) Number of nuclei present after 1.2 min:

Decay constant, $lambda equals fraction numerator 0.693 over denominator t subscript bevelled 1 half end subscript end fraction equals fraction numerator 0.693 over denominator 30.8 end fraction s e c to the power of negative 1 end exponent$

time, t $equals 1.2 m i n equals 72 space s e c$

So, number of nuclei after 1.2 min,

$N equals N subscript 0 e to the power of negative lambda t end exponent rightwards double arrow N equals 30.120 cross times 10 to the power of 15 cross times e to the power of negative fraction numerator 0.693 over denominator 30.8 end fraction cross times 72 end exponent rightwards double arrow N equals 5.96 cross times 10 to the power of 15$

(3) Activity after 1.2 min,

$A equals lambda N rightwards double arrow A equals fraction numerator 0.693 over denominator 30.8 end fraction cross times 5.96 cross times 10 to the power of 15 rightwards double arrow A equals 13.41 cross times 10 to the power of 13 space s e c to the power of negative 1 end exponent$

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