Home Page Textbooks Physics Page 925: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 925: Lesson Check

Exercise 16

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For heavier nucleus, some of the mass will be converted to the energy after decay. Hence the total mass of the nucleus after decay will be less than before decay.

Exercise 17

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Suppose we have, at the beginning, a radioactive element. In some cases, that element will directly decay into a stable element (isotope). But, in some other cases, the element will undergo numerous ‘transformations’, ie. it will decay into different isotopes several times, until it reaches the final stable isotope. In this way a radioactive decay series is produced.

Result

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A radioactive decay series is a process in which a radioactive element undergoes several decays, ie. it decays into several different isotopes until it reaches a stable one.

Exercise 18

Solution 1

Solution 2

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When a particle encounters its antiparticle counterpart, they will annihilate each other and disappear – but this process creates energy and in some cases, form other particles as well.

Result

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Particles and antiparticles annihilate each other and creates a burst of energy in the process

Step 1

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They will annihilate and produce energy in the form of gamma ray.

Exercise 19

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begin{tabular}{|c|c|c|c|}
hline
& atomic number & neutron number & mass numbertabularnewline
hline
hline
$alpha$ & decreses by 2 & decreases by 2 & decreases by 4tabularnewline
hline
$beta$ & increases by 1 & unchanged & unchangedtabularnewline
hline
$gamma$ & unchanged & unchanged & unchangedtabularnewline
hline
end{tabular}

Exercise 20

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When a nucleus emits an alpha particle ($^4_2$He), it looses two protons and two neutrons.

Suppose $Z$ is the atomic number of parent nucleus then the atomic number of daughter nucleus is $Z-2$.

If $N$ is the neutron number of parent nucleus, the neutron number of daughter nucleus is $N-2$.

The mass number of the daughter nucleus is $(Z-2)+(N-2) = Z -2 + N – 2 = Z+N -4 = A-4$.

where $A = Z+N$ is the mass number of the parent nucleus.

Result

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(a) $A-4$

(b) $Z-2$

Exercise 21

Step 1

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In beta decay, one neutron is converted into a proton and an electron.

Suppose $Z$ is the atomic number of the parent nucleus then the atomic number of daughter nucleus is $Z+1$.

If$N$ is the neutron number of parent nucleus then the neutron number of daughter nucleus is $N-1$.

The mass number of the daughter nucleus is $(Z+1)+(N-1) = Z+1 +N-1 = Z+N =A$

where $A = Z+N$ is the mass number of the parent nucleus. So the mass number of the daughter nucleus is same as parent nucleus.

Result

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(a) $A$

(b) $Z+1$

Exercise 22

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In alpha decay, the parent nucleus looses two neutrons and two protons. So the atomic number of the daughter nucleus is $Z-2$ and its mass number is $A-4$.

For the nucleus $^{227}_{89}$Ac, $Z=89$ and $A=227$.

When it undergoes alpha decay, the atomic number of daughter nucleus is $89-2=87$ and the mass number is $227-4=223$.

Francium (Fr) has atomic number 87. So the daughter nucleus is $^{223}_{87}$Fr.

$^{227}_{89}$Ac $rightarrow$ $^{223}_{87}$Fr $+$ $^4_2$H $+$ energy

Result

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$^{223}_{87}$Fr

Exercise 23

Step 1

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In beta decay, a neutron is converted into a proton and an electron.

So the atomic number of the daughter nuclei increases by 1 and its mass number remains same as the atomic number.

When $^{215}_{83}$Bi undergoes beta decay, the atomic number of the daughter nucleus is $83+1 = 84$ and the mass number is 215 (same as parent nucleus).

Polonium (Po) is the element with atomic number 84.

So the daughter nucleus is $^{215}_{84}$Po.

$^{215}_{83}$Bi $rightarrow$ $^{215}_{84}$Po $+$ e$^{-}$ $+$ $bar{gamma_e}$ $+$ energy

Result

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$^{215}_{84}$Po

Exercise 24

Step 1

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(a) In alpha decay, the parent nucleus looses two neutrons and two protons. So the atomic number of the daughter nucleus is $Z-2$ and its mass number is $A-4$.

For the nucleus $^{219}_{86}$Rn, $Z=86$ and $A=219$.

When it undergoes alpha decay, the atomic number of daughter nucleus is $86-2=84$ and the mass number is $219-4=215$.

Polonium (Po) has atomic number 84. So the daughter nucleus is $^{215}_{84}$Po.

$^{219}_{86}$Rn $rightarrow$ $^{215}_{84}$Po $+$ $^4_2$H $+$ energy

Step 2

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(b) In beta decay, a neutron is converted into a proton and an electron.

So the atomic number of the daughter nuclei increases by 1 and its mass number remains same as the atomic number.

When $^{219}_{86}$Rn undergoes beta decay, the atomic number of the daughter nucleus is $86+1 = 87$ and the mass number is 219 (same as parent nucleus).

Francium (Fr) is the element with atomic number 87.

So the daughter nucleus is $^{219}_{87}$Fr.

$^{219}_{86}$Rn $rightarrow$ $^{219}_{87}$Fr $+$ e$^{-}$ $+$ $bar{gamma_e}$ $+$ energy

Result

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(a) $^{215}_{84}$Po

(b) $^{219}_{87}$Fr

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