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51.

Write two units of radioactivity. How are they interrelated?

Answer»

i. The unit of radioactivity is curie (Ci).

1 Ci = 3.7 × 1010 dps

ii. Other unit of radioactivity is Becquerel (Bq).

1 Bq = 1 dps

Thus, 

1 Ci = 3.7 × 1010 dps 

= 3.7 × 1010 Bq
Discussion
52.

The half-life of \(^{32}\)P is 14.26 d. What percentage of \(^{32}\)P sample will remain after 40 d?

Answer»

Given: t1/2 = 14.26 d, 

N0 = 100, 

t = 40 d 

To find: Percentage of \(^{32}\)P sample remaining after 40 d

Formula:

i. t1/2 = \(\frac{0.693}{\lambda}\)

ii. \(\lambda\) = \(\frac{2.303}{t}\) log10 \((\frac{N_0}{N})\)

Calculation: \(\lambda\) = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{14.26\,d}\) = 0.0486 d-1

Now, \(\lambda\) = \(\frac{2.303}{t}\) log10 \((\frac{N_0}{N})\)

Hence,  log10 \(\frac{N_0}{N}\) = \(\frac{\lambda t}{2.303}\)

= \(\frac{0.0486\,d^{-1}\times 40\,d}{2.303}\) = 0.8441

Taking antilog of both sides we get,

\(\frac{N_0}{N}\) = antilog (0.8441) = 6.984

∴ \(\frac{100}{N}\) = 6.984 or N = \(\frac{100}{6.984}\) = 14.32%

% \(^{32}P\) that remains after 40 d is 14.32%.
Discussion
53.

The half-life of \(^{209}\)Po is 102 y. How much of 1 mg sample of polonium decays in 62 y?

Answer»

Given: t1/2 = 102y,

t = 62 y, 

N0 = 1 mg 

To find: Amount of polonium that decayed in 62 y

Formula: 

i. t1/2 = \(\frac{0.693}{\lambda}\)
ii.\(\lambda\) = \(\frac{2.303}t\) log10 \((\frac{N_0}N)\)

Calculation:  \(\lambda\) = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{102y}\) = 6.794 x 10-3 y-1

\(\lambda\) = \(\frac{2.303}{t}\)  log10 \((\frac{N_0}N)\)

Hence,  log10 \((\frac{N_0}N)\) = \(\frac{\lambda t}{2.303}\)

= \(\frac{6.794\times 10^{-3}y^{-1}\times 62\,y}{2.303}\) = 0.1829

Taking antilog of both sides we get, \(\frac{N_0}N\) = antilog (0.1829) = 1.524

N = \(\frac{N_0}{1.524}\) = \(\frac{1\,mg}{1.524}\) = 0.656 mg

N is the amount that remains after 62 y. Hence, the amount decayed in 62 y = 1 mg – 0.656 mg = 0.344 mg

The amount decayed in 62 y is 0.344 mg
Discussion
54.

\(^{218}PO\) decays initially at a rate of 816 dps. The rate falls to 408 dps after 24 min. Calculate the decay constant.

Answer»

Given: \((\frac{-dN_0}{dt})\) = 816 dps, \((\frac{-dN}{dt})\) = 408 dps, t = 24 min

To find: Decay constant \((\lambda)\)

Formula: \(\lambda\) = \(\frac{2.303}t\) log10 \((\frac{N_0}N)\)

Calculation: \(\lambda\) = \(\frac{2.303}t\)  log10 \((\frac{N_0}N)\)

\(\lambda\) = \(\frac{2.303}{24}\) log10 \((\frac{816}{408})\)

= 0.0288 = 0.029 min-1 

Decay constant \((\lambda)\) will be 0.029 min-1 
Discussion