(a) The half life period of radium is 1590 yrs. After how many years will one gram of the pure element,
(i)be reduced to one centigram, (ii) lose one centigram. (b) The half life of radon is 3.8 days. After how many days will only one twentieth of radon sample be left over?
(c) 1 gm of radioactive substance takes 50 sec. to lose 1 centigram. Find its half life period?

Answer

Let's solve the question step by step.

### Part (a)

**Given:**
- Half-life period of radium (T_half) = 1590 years
- Initial mass of radium (N0) = 1 gram
- Final mass for (i) = 0.01 grams (1 centigram)
- Final mass for (ii) = 0.99 grams (after losing 1 centigram)

#### (i) Time to reduce 1 gram to 1 centigram

1. **Determine the ratio (α)**:
\[ \alpha = \frac{N}{N_0} = \frac{0.01 \text{ g}}{1 \text{ g}} = 0.01 \]

2. **Use the formula for time (t)**:
\[ t = \frac{\ln\left(\frac{1}{\alpha}\right)}{\lambda} \]
where \(\lambda = \frac{0.693}{T_{half}}\).

3. **Calculate λ**:
\[ \lambda = \frac{0.693}{1590} \approx 0.000436 \]

4. **Substitute α and λ into the time formula**:
\[ t = \frac{\ln\left(\frac{1}{0.01}\right)}{0.000436} = \frac{\ln(100)}{0.000436} \approx \frac{4.605}{0.000436} \approx 10567.9 \text{ years} \]

#### (ii) Time to lose 1 centigram

1. **Determine the ratio (α)**:
\[ \alpha = \frac{N}{N_0} = \frac{0.99 \text{ g}}{1 \text{ g}} = 0.99 \]

2. **Use the same time formula**:
\[ t = \frac{\ln\left(\frac{1}{0.99}\right)}{\lambda} \]

3. **Substitute λ**:
\[ t = \frac{\ln\left(\frac{1}{0.99}\right)}{0.000436} = \frac{\ln(1.0101)}{0.000436} \approx \frac{0.01005}{0.000436} \approx 23.06 \text{ years} \]

### Part (b)

**Given:**
- Half-life of radon (T_half) = 3.8 days
- We want to find the time when only \(\frac{1}{20}\) of the sample is left.

1. **Determine the ratio (α)**:
\[ \alpha = \frac{1}{20} \implies N = \frac{N_0}{20} \]

2. **Use the time formula**:
\[ t = \frac{\ln\left(\frac{1}{\alpha}\right)}{\lambda} \]

3. **Calculate λ**:
\[ \lambda = \frac{0.693}{3.8} \approx 0.1824 \]

4. **Substitute α and λ**:
\[ t = \frac{\ln(20)}{0.1824} \approx \frac{2.9957}{0.1824} \approx 16.43 \text{ days} \]

### Part (c)

**Given:**
- Initial mass = 1 gram
- Takes 50 seconds to lose 1 centigram.

1. **Determine the remaining mass**:
\[ N = 1 \text{ g} - 0.01 \text{ g} = 0.99 \text{ g} \]

2. **Determine the ratio (α)**:
\[ \alpha = \frac{N}{N_0} = \frac{0.99 \text{ g}}{1 \text{ g}} = 0.99 \]

3. **Use the time formula**:
\[ t = 50 \text{ seconds} \]

4. **Rearranging for T_half**:
\[ T_{half} = \frac{0.693}{\ln\left(\frac{1}{\alpha}\right)} \cdot t \]

5. **Substituting values**:
\[ T_{half} = \frac{0.693}{\ln\left(\frac{1}{0.99}\right)} \cdot 50 \approx \frac{0.693}{0.01005} \cdot 50 \approx 3447.64 \text{ seconds} \]

### Summary of Answers:
- (a)(i) 10567.9 years
- (a)(ii) 23.06 years
- (b) 16.43 days
- (c) 3447.64 seconds