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A) A Beta particle is an electron (^0-1e) or positron (^0+1e) emitted from the nucleus of an atom during radioactive decay. In the context of typical beta decay, it refers to an electr

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b
223
101
c(ii)

13.3 years13.3 \, \text{years}

Here are the solutions to the radioactivity questions:

a) A Beta particle is an electron (−10e^0_{-1}e) or positron (+10e^0_{+1}e) emitted from the nucleus of an atom during radioactive decay. In the context of typical beta decay, it refers to an electron.

b) The initial radioactive element is 98223X^{223}_{98}\text{X}.
A Beta particle emission (−10e^0_{-1}e) changes the nuclide as follows:

  • The nucleon number (mass number, A) remains unchanged.
  • The proton number (atomic number, Z) increases by 1.

The element emits a total of 1+2=31 + 2 = 3 Beta particles.

After 3 Beta emissions:

  • Nucleon number: 223+(3×0)=223223 + (3 \times 0) = 223
  • Proton number: 98+(3×1)=98+3=10198 + (3 \times 1) = 98 + 3 = 101

The nucleon number of the remaining nuclide is 223\boxed{223}.
The proton number of the remaining nuclide is 101\boxed{101}.

c) A 400g radioactive sample has a half-life of 4 years.

i) Plot a graph to show this decay curve after the period of 32 years.
To plot the decay curve, we calculate the mass remaining after each half-life.
Initial mass (N0N_0) = 400 g
Half-life (T1/2T_{1/2}) = 4 years
Total time = 32 years

The number of half-lives (nn) is Total timeHalf-life=32 years4 years=8\frac{\text{Total time}}{\text{Half-life}} = \frac{32 \, \text{years}}{4 \, \text{years}} = 8.

The mass remaining (NN) after nn half-lives is given by N=N0(12)nN = N_0 \left(\frac{1}{2}\right)^n.

Here are the data points for the decay curve:

Time (years)Number of half-lives (nn)Mass remaining (g) (400×(1/2)n400 \times (1/2)^n)
00400
41400×12=200400 \times \frac{1}{2} = 200
82200×12=100200 \times \frac{1}{2} = 100
123100×12=50100 \times \frac{1}{2} = 50
16450×12=2550 \times \frac{1}{2} = 25
20525×12=12.525 \times \frac{1}{2} = 12.5
24612.5×12=6.2512.5 \times \frac{1}{2} = 6.25
2876.25×12=3.1256.25 \times \frac{1}{2} = 3.125
3283.125×12=1.56253.125 \times \frac{1}{2} = 1.5625

To plot the graph:

  • Draw a set of axes on graph paper.
  • Label the x-axis "Time (years)" from 0 to 32 years.
  • Label the y-axis "Mass remaining (g)" from 0 to 400 g.
  • Plot the points from the table above.
  • Draw a smooth curve connecting these points. The curve should show an exponential decrease in mass over time.

ii) What period of time would it take for the sample to reduce to 40g?
Initial mass (N0N_0) = 400 g
Final mass (NN) = 40 g
Half-life (T1/2T_{1/2}) = 4 years

We use the formula N=N0(12)nN = N_0 \left(\frac{1}{2}\right)^n, where nn is the number of half-lives.

40=400(12)n40 = 400 \left(\frac{1}{2}\right)^n 40400=(12)n\frac{40}{400} = \left(\frac{1}{2}\right)^n 0.1=(12)n0.1 = \left(\frac{1}{2}\right)^n

To solve for nn, take the logarithm of both sides:

log⁡(0.1)=nlog⁡(0.5)\log(0.1) = n \log(0.5) n=log⁡(0.1)log⁡(0.5)n = \frac{\log(0.1)}{\log(0.5)} n≈−1−0.30103n \approx \frac{-1}{-0.30103} n≈3.3219n \approx 3.3219

Now, calculate the total time (TT):

T=n×T1/2T = n \times T_{1/2} T=3.3219×4 yearsT = 3.3219 \times 4 \, \text{years} T≈13.2876 yearsT \approx 13.2876 \, \text{years}

Rounding to 3 significant figures:
The period of time is approximately 13.3 years\boxed{13.3 \, \text{years}}.

iii) Name a source of gamma radiation.
A common source of gamma radiation is Cobalt-60 (or Caesium-137).

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