Home Page Textbooks Physics Page 917: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 917: Lesson Check

Exercise 3

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The $textbf{atomic number}$ (Z) describes how many protons are in the nucleus of a certain element, the $textbf{neutron number}$ (N) describes how many neutrons are in a certain element, while the $textbf{mass number}$ (A) describes the number of nucleons (combined number of protons and neutrons in the nucleus). The mass number can be represented using the atomic and neutron number, moreover, these three quantities can help identify an element (or its isotopes) such that:

A = Z + N

The nucleus of an element X, with atomic number Z, and mass number A can be expressed as:
$_{Z}^{A} X$

Result

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The nucleus of an element X, with atomic number Z, and mass number A can be expressed as:
$_{Z}^{A} X$

The atomic number (Z), mass number (A), and neutron number (N) can be used to identify an element (or its isotopes). Check explanation

Exercise 4

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In the frame of reference in which the particle of mass $m$ is at rest, its energy $E$ is given by

$$
begin{align*}
E=mc^2
end{align*}
$$

where $c$ is the speed of light in vacuum. This equation is sometimes referred to as the energy-mass equivalence.

Result

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$$
E=mc^2
$$

Exercise 5

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The nucleus is combined of protons and electrons. If the only force acting in the nucleus was the electromagnetic force, the nucleus would fall apart; this is because the electromagnetic force does not act upon neutrons and the repulsive force between the protons would tend to break apart the nucleus.

Now, we know that there exists a force called the strong nuclear force. It is responsible for the interraction between the protons and neutrons in the nucleus, and thus keeping the nucleus stable.

The strong nuclear force only acts in the nucleus. Because it is of very short range, it does not act on the electrons which are not a part of the nucleus.

Result

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Please click on the solution.

Exercise 6

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If a nucleus gains a neutron, then the neutron number increases by one and the mass number which is sum of neutron number and atomic number increases by one.

The atomic number is the number of protons and so it remains same.

Result

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See Explanation.

Exercise 7

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Mass number, $A = 182$.

Atomic number, $Z = 74$.

So the neutron number, $N = A – Z = 182 – 74 = 108$.

If two neutrons are lost then the neutron number becomes

$N= 108 – 2 = 106$.

The atomic number $Z$ remains same because it represents the total number of protons.

Mass number is $A = Z + N = 74 + 106 = 180$.

Result

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(a) $A = 180$
(b) $Z = 74$
(c) $N = 106$

Exercise 8

Solution 1

Solution 2

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Given:

Number of protons = 25

Number of neutrons = 30

Atomic number (Z) = number of protons

Neutron number (N) = number of neutrons

Mass number (A) = number of nucleons (protons + neutrons)

To compute for the mass number we have the formula: A = Z + N

$$
begin{align*}
A &= 25 + 30 \
&= 50
end{align*}
$$

The nucleus of an element X can be written as: $_{Z}^{A} X$

Therefore, the symbol for the nucleus of this element is: $boxed{_{25}^{55} X}$ or if you look it up in the periodic table, the element here is Manganese and its symbol for its nucleus is $_{25}^{55} Mn$

Result

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$_{25}^{55} X$ where X is the symbol of the element

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Number of protons is 25. So the atomic number $Z = 25$.

Number of neutrons is 30. So the neutron number $N = 30$.

Mass number $A = Z + N = 25 + 30 = 55$.

The nucleus of an element X with atomic number $Z$ and mass number $A$ is denoted as $^A_Z$X.

So the nucleus with 25 protons, 30 neutrons and so 55 nucleons is denoted as

$^{55}_{25}$X.

Result

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See Solution.

Exercise 9

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In a nuclear reaction, the atomic number $Z$ and mass number $A$ are conserved.

The total atomic number on the left side of the equation is $0 + 92 = 92$.

Let the atomic number of the unknown nucleus is $x$. Then the total atomic number on the right side of the equation is $56 + x + 0 + 0 = 56 +x$.

Since the atomic number has to be same on both sides we have

$92 = 56 + x$

So, $x = 92 – 56 = 36$

Kr is the element with atomic number 36.

Let us find the mass number:

The total mass number on left side is $1 + 235 = 236$.

Let $y$ be the mass number of nucleus of Kr. Then the total mass number on right side is $144 + y + 1 + 1 = 146 +y$.

Conservation of mass number leads to the equation,

$236 = 146 + y$

So, $y = 236 – 146 = 90$.

So the missing nucleus is $^{90}_{36}$Kr.

Result

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$^{90}_{36}$Kr

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