Some amount of a radioactice substance (half-life =10 days ) is spread inside a room and consequently the level of radiation become `50` times the permissible level for normal occupancy of the room. After how many days will the room be safe for occupation?.

Options

A. 20 days
B. 34.8 days
C. 56.4 days
D. 62.9 days

Answer

To solve the problem step by step, we will use the concept of radioactive decay and the half-life of the substance.

### Step 1: Understand the problem
We have a radioactive substance with a half-life of 10 days. The radiation level in the room is currently 50 times the permissible level. We need to find out after how many days the radiation level will drop to a safe level (1 times the permissible level).

### Step 2: Define the variables
Let:
- \( N_0 \) = initial radiation level (which is 50 times the permissible level)
- \( N(t) \) = radiation level after time \( t \)
- \( N(t) \) should equal the permissible level, which we can denote as \( 1x \).

### Step 3: Use the radioactive decay formula
The radioactive decay can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 4: Calculate the decay constant
The decay constant \( \lambda \) can be calculated using the half-life (\( t_{1/2} \)):
\[ \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{10 \text{ days}} \]

### Step 5: Set up the equation
We want to find \( t \) when \( N(t) = 1x \):
\[ 1x = 50x \cdot e^{-\lambda t} \]
Dividing both sides by \( x \):
\[ 1 = 50 \cdot e^{-\lambda t} \]
Rearranging gives:
\[ e^{-\lambda t} = \frac{1}{50} \]

### Step 6: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ -\lambda t = \ln\left(\frac{1}{50}\right) \]
This simplifies to:
\[ t = -\frac{\ln\left(\frac{1}{50}\right)}{\lambda} \]

### Step 7: Substitute the decay constant
Substituting \( \lambda \):
\[ t = -\frac{\ln\left(\frac{1}{50}\right)}{\frac{\ln(2)}{10}} \]
This can be rewritten as:
\[ t = -10 \cdot \frac{\ln\left(\frac{1}{50}\right)}{\ln(2)} \]

### Step 8: Calculate the values
Now, we calculate:
- \( \ln(50) \approx 3.912 \)
- \( \ln(1) = 0 \)

Thus:
\[ \ln\left(\frac{1}{50}\right) = -\ln(50) \approx -3.912 \]
Now substituting:
\[ t = -10 \cdot \frac{-3.912}{\ln(2)} \approx -10 \cdot \frac{-3.912}{0.693} \approx 56.4 \text{ days} \]

### Final Answer
The room will be safe for occupation after approximately **56.4 days**.

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