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charle [14.2K]
3 years ago
12

Determine which decay process is being described in each case. the atomic number deceases by one, and the atomic mass is unchang

ed. the atomic number decreases by two, and the atomic mass decreases by four. the atomic number increases by one, and the atomic mass remains unchanged.
Chemistry
2 answers:
Sav [38]3 years ago
8 0
1. The atomic number deceases by one and the atomic mass is unchanged - β⁺/positron emission. In this a proton is converted to a neutron, hence no net change in mass. Since a proton is converted to neutron, the daughter nuclide's proton number /atomic number decreases by 1. Then atomic number decreases by 1 and no change in mass.

2.the atomic number decreases by two, and the atomic mass decreases by four - alpha decay / ⁴₂α. alpha particles have 2 neutrons and 2 protons. Since 2 protons and 2 neutrons are emitted, the mass of the nuclide decreases by 4. Since 2 protons have been emitted, this results in atomic number decreasing  by 2.

3.<span>the atomic number increases by one, and the atomic mass remains unchanged. - </span>β⁻ beta decay. In this type of decay - beta decay , a neutron is converted to a proton, therefore no net change in mass. Since a proton is formed, atomic number increases by 1. Therefore atomic number increases by 1 and no change in mass.
swat323 years ago
8 0

Determine which decay process is being described in each case.

The atomic number deceases by one, and the atomic mass is unchanged.      

beta-plus decay

The atomic number decreases by two, and the atomic mass decreases by four.      

alpha-decay

The atomic number increases by one, and the atomic mass remains unchanged.      

beta-minus decay

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Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

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You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

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For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

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Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

  • 0 = Vo - gt

   

  • Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

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