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gavmur [86]
3 years ago
9

After a nucleus undergoes radioactive decay, its new mass number is:

Physics
2 answers:
Ivanshal [37]3 years ago
7 0
Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.
lapo4ka [179]3 years ago
7 0

<u>Answer:</u> The new mass number is never more than its original mass number

<u>Explanation:</u>

Radioactive decay is defined as the process in which an unstable nuclei breaks down into stable nuclei via various methods.

An isotope undergoes a radioactive decay to attain stability.

There are many decay processes by which a parent nucleus can undergo decay. They are:

Alpha decay is defined as the decay process in which alpha particle is released. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units and a mass of 4 units.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

Beta decay is defined as the decay process in which a neutron gets converted to a proton and an electron.  In this decay process, beta particle is emitted. The emitted particle carries a charge of -1 units and has a mass of 0 units. The released beta particle is also known as electron.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

Gamma decay is defined as the decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. This decay releases \gamma -radiations. This process does not change the mass number.

_A^Z\textrm{X}^*\rightarrow _A^Z\textrm{X}+_0^0\gamma

For every decay process, the mass number will either remain same or the mass number of new isotope will be less than the parent isotope but not more than the original mass number.

Hence, the new mass number is never more than its original mass number

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A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
Which of these accurately describes the products of a reaction?
stiv31 [10]
The one that accurately describes the products of a reaction is : B. new substances that are present at the end of a reaction
For example the process of photosynthesis transform CO2 and other nutrients into O2 and H2O

hope this helps
3 0
3 years ago
Read 2 more answers
The answer would be ?
Cerrena [4.2K]

Answer:

D. demand; increased

Explanation:

Demand is how much people want it.

4 0
3 years ago
A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

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3 years ago
Name the Organs and functions of digestive system
Alexxx [7]
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6 0
3 years ago
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