Home Page Textbooks Physics: Principles and Problems Page 831: Standardized Test Practice

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 831: Standardized Test Practice

Exercise 1

Step 1

1 of 2

In this problem we are given an isotope $^{60}_{28}$Ni for which we want to know the number of protons, neutrons and electrons. Since by definition we have that $^A_Z X$ where

$$
A=n_p+n_n
$$

$$
Z=n_p=n_e
$$

we conclude that the correct answer is A.

$$
n_p=28;, n_n=32;, n_e=28
$$

Result

2 of 2

$$
textrm{A) }n_p=28;, n_n=32;, n_e=28
$$

Exercise 2

Step 1

1 of 2

By definition a type of reaction that is given by the following formula

$$
^{A}_{Z}Xrightarrow ^{A}_{Z-1}Y+e^-barnu
$$

is called $beta$-decay so here the correct answer is B.

Result

2 of 2

B. $beta$-decay

Exercise 3

Step 1

1 of 4

In this problem, we are asked to write the nuclear equation for $alpha$-decay of the polonium isotope $^{210}_{84}$Po.

Step 2

2 of 4

In order to do so, we are going to write the general $alpha$-decay equation which is given as

$$
^{A}_ZXrightarrow^{A-4}_{Z-2}Y+^4_2alpha
$$

Step 3

3 of 4

From the equation above we see that $Z-2=82$ and $A-4=206$ so the correct answer is A, $^{206}_{82}$Pb.

Result

4 of 4

A) $^{206}_{82}$Pb.

Exercise 4

Step 1

1 of 4

In this problem we are given a sample of the radioactive iodine with a given radioactivity and half-life. We want to know its activity after 16 days.

Step 2

2 of 4

In order to answer this question, we are going to use the half-life relation which tells us that

$$
A=A_0cdot 2^{-frac{T}{T_{1/2}}}
$$

Now, we can substitute the given values to have that after 16 days

Step 3

3 of 4

[A=2.5times 10^8cdot 2^{-frac{16}{8}}=6.25times 10^7textrm{ Bq}]
center{so the final answer is B.}

Result

4 of 4

$$
textrm{B)} A=6.25times 10^7textrm{ Bq}
$$

Exercise 5

Step 1

1 of 3

In this problem we are given a reaction given by the following equation

$$
^1_0textrm{n}+{^{14}_{7}textrm{N}}rightarrow{^{14}_{6}textrm{C}}+^{textrm{A}}_{textrm{Z}}textrm{X}
$$

for which we should find the unknown isotpe.

Step 2

2 of 3

Now, we can find the relevant parameters by taking the following relations

$$
A=1+14-14=1
$$

$$
Z=7-6=1
$$

so our mysterious element is $^1_1$H, i.e. the choice A.

Result

3 of 3

A) $^1_1$H

Exercise 6

Step 1

1 of 2

The only decay process in which no particles are ejected from the nucleus whatsoever is gamma decay where a high-energy photon is ejected instead of the particles so the answer is D.

Result

2 of 2

D) gamma decay

Exercise 7

Step 1

1 of 3

In this problem we are given a polonium sample with a given half-life. We want to know how much of that sample remains after four years.

Step 2

2 of 3

In order to solve this problem, we are going to use the half-life formula which tells us that

$$
m=m_0cdot2^{-frac{T}{T_{1/2}}}=2.34times 2^{-frac{4times 365}{138}}=1.51times 10^{-3}textrm{ kg}
$$

so the answer is C.

Result

3 of 3

$$
textrm{C) }m=1.51times 10^{-3}textrm{ kg}
$$

Exercise 8

Step 1

1 of 2

The energy equivalent will be equal to the sum of the energy equivalents of the positron and electron i.e. 1.02 MeV so the correct answer is B.

Result

2 of 2

B) 1.02 MeV

Exercise 9

Step 1

1 of 2

The reason for gamma rays not to leave the trace is the fact that the bubble chamber works based on repulsive electrostatic interactions and gamma rays are neutral so the answer is D.

Result

2 of 2

The answer is D.

Exercise 10

Step 1

1 of 3

In this problem we are given the energy released per nucleus during the fission of U-235 and the energy released per ton of TNT. We want to know how many nuclei of U-235 do we need for 20,000 tons of TNT.

Step 2

2 of 3

In order to solve this problem, we are simply going to multiply the number of tons by energy released per ton and divide by the energy released per nucleus during the fission reaction.

$$
N_U=frac{Ntimes E_{TNT}}{E_{U}}=frac{20times 10^3times 4times 10^9}{3.2times 10^{-11}}=boxed{25times 10^{23}}
$$

Result

3 of 3

$$
N_U=25times 10^{23}
$$

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