To develop this problem it is necessary to apply the concepts related to Sound Intensity.
By definition the intensity is given by the equation
Where,
I = Intensity of Sound
= Intensity of Reference
At this case we have that 15 engines produces 15 times the reference intensity, that is
And the total mutual intensity is 100 dB, so we should
Therefore each one of these engines produce D. 88dB.
A) Francium-223
In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.
This means that in the decay:
- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units
- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units
In this example, the original nucleus is Ac (Actinium), with
Z = 89
A = 227
After the decay, it must be
Z - 2 = 89 - 2 = 87
A - 4 = 227 - 4 = 223
We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).
B) Polonium-211
In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.
Therefore, in this process:
- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit
- The original nucleus does not lose/gain nucleons, so its mass number A remains the same
In this example, the original nucleus is Bi (bismuth)-211, with
Z = 83
A = 211
So After the decay, it will be
Z + 1 = 83 + 1 = 84
A = 211
So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.
C) Neon-22
In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.
Therefore, in this process:
- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit
- The original nucleus does not lose/gain nucleons, so its mass number A remains the same
In this example, the original nucleus is Na (sodium)-22, with
Z = 11
A = 22
So After the decay, it will be
Z - 1 = 11 - 1 = 10
A = 22
So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.
D) Technetium-98
In a gamma decay, an unstable nucleus emits a gamma ray:
In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:
- The atomic number Z remains constant
- The mass number A remains constant
In this example, we have a nucleus of Tc (Technetium)-98, with
Z = 43
A = 98
These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.