Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as
`-(dN)/(dt)=lambdaN` Where `lambda=` decay constant, N= number of nuclei at time t, `N_(0)` =intial no. of nuclei. The above equation after integration can be represented as
`lambda=(2.303)/(t)log((N_(0))/(N))`
Calculate the half-life period of a radioactive element which remains only `1//16` of its original amount in `4740` years:

Options

A. 1185 years
B. 2370 years
C. `52.5` years
D. none of these

Answer

To calculate the half-life period of a radioactive element that remains only \( \frac{1}{16} \) of its original amount in \( 4740 \) years, we can follow these steps:

### Step 1: Understand the relationship between the original amount and the remaining amount.
The problem states that the remaining amount is \( \frac{1}{16} \) of the original amount. This means that the original amount has undergone several half-lives to reach this point.

### Step 2: Determine how many half-lives it takes to reach \( \frac{1}{16} \).
Since \( \frac{1}{16} \) can be expressed as \( \left( \frac{1}{2} \right)^4 \), it indicates that it takes 4 half-lives to reduce the original amount to \( \frac{1}{16} \).

### Step 3: Relate the total time to the number of half-lives.
The total time taken to reach \( \frac{1}{16} \) of the original amount is given as \( 4740 \) years. Since we determined that this corresponds to 4 half-lives, we can set up the equation:
\[ \text{Total time} = \text{Number of half-lives} \times \text{Half-life period} \]
\[ 4740 \text{ years} = 4 \times t_{1/2} \]

### Step 4: Solve for the half-life period \( t_{1/2} \).
To find the half-life period, rearrange the equation:
\[ t_{1/2} = \frac{4740 \text{ years}}{4} \]
Calculating this gives:
\[ t_{1/2} = 1185 \text{ years} \]

### Conclusion:
The half-life period of the radioactive element is \( 1185 \) years.