A radio nuclide consists of two isotopes. One of the isotopes decays by `alpha`-emission and other by `beta`-emission with half-lives `T_1=405s` and `T_2=1620s`, respectively. At `t=0`, probabilities of getting `alpha` and `beta`-particles from the radio nuclide are equal . Calculate their respective probabilities at `t=1620s`. If at `t=0`, total number of nuclei in the radio nuclide are `N_0`. Calculate the time t when total number of nuclei remained undecayed becomes equal to `N_0//2`.
`log_(10)2=0.3010`, `log_(10)5.94=0.7742` and `x^4+4x-2.5=0, `x=0.594`

Explanation

To solve the problem step by step, we will break it down into two parts as described in the question.

### Part 1: Calculate the probabilities of getting alpha and beta particles at \( t = 1620 \, s \)

1. **Understanding Initial Conditions**:
At \( t = 0 \), the probabilities of getting alpha and beta particles are equal. This means that the initial activities of both isotopes are the same, denoted as \( R_0 \).

2. **Calculating Activities**:
The activity of a radioactive substance is given by:
\[ R = R_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
For the alpha-emitting isotope (half-life \( T_1 = 405 \, s \)):
\[ R_1 = R_0 \left( \frac{1}{2} \right)^{\frac{1620}{405}} = R_0 \left( \frac{1}{2} \right)^{4} = \frac{R_0}{16} \]

For the beta-emitting isotope (half-life \( T_2 = 1620 \, s \)):
\[ R_2 = R_0 \left( \frac{1}{2} \right)^{\frac{1620}{1620}} = R_0 \left( \frac{1}{2} \right)^{1} = \frac{R_0}{2} \]

3. **Total Activity**:
The total activity \( R \) at \( t = 1620 \, s \) is:
\[ R = R_1 + R_2 = \frac{R_0}{16} + \frac{R_0}{2} \]
To combine these fractions, we can express \( \frac{R_0}{2} \) as \( \frac{8R_0}{16} \):
\[ R = \frac{R_0}{16} + \frac{8R_0}{16} = \frac{9R_0}{16} \]

4. **Calculating Probabilities**:
- Probability of getting an alpha particle:
\[ P_{\alpha} = \frac{R_1}{R} = \frac{\frac{R_0}{16}}{\frac{9R_0}{16}} = \frac{1}{9} \]
- Probability of getting a beta particle:
\[ P_{\beta} = \frac{R_2}{R} = \frac{\frac{R_0}{2}}{\frac{9R_0}{16}} = \frac{8}{9} \]

### Part 2: Calculate time \( t \) when total number of undecayed nuclei becomes \( \frac{N_0}{2} \)

1. **Initial Nuclei Distribution**:
Let \( N_0 \) be the total number of nuclei at \( t = 0 \). From the equal probabilities of decay, we can say:
\[ N_{1,0} = \frac{N_0}{5}, \quad N_{2,0} = \frac{4N_0}{5} \]

2. **Undecayed Nuclei at Time \( t \)**:
The number of undecayed nuclei after time \( t \) for each isotope is given by:
\[ N_1 = N_{1,0} \left( \frac{1}{2} \right)^{\frac{t}{T_1}} = \frac{N_0}{5} \left( \frac{1}{2} \right)^{\frac{t}{405}} \]
\[ N_2 = N_{2,0} \left( \frac{1}{2} \right)^{\frac{t}{T_2}} = \frac{4N_0}{5} \left( \frac{1}{2} \right)^{\frac{t}{1620}} \]

3. **Total Undecayed Nuclei**:
The total number of undecayed nuclei is:
\[ N = N_1 + N_2 = \frac{N_0}{5} \left( \frac{1}{2} \right)^{\frac{t}{405}} + \frac{4N_0}{5} \left( \frac{1}{2} \right)^{\frac{t}{1620}} \]
We want to find \( t \) when \( N = \frac{N_0}{2} \):
\[ \frac{N_0}{5} \left( \frac{1}{2} \right)^{\frac{t}{405}} + \frac{4N_0}{5} \left( \frac{1}{2} \right)^{\frac{t}{1620}} = \frac{N_0}{2} \]

4. **Simplifying the Equation**:
Dividing through by \( N_0 \):
\[ \frac{1}{5} \left( \frac{1}{2} \right)^{\frac{t}{405}} + \frac{4}{5} \left( \frac{1}{2} \right)^{\frac{t}{1620}} = \frac{1}{2} \]
Let \( x = \left( \frac{1}{2} \right)^{\frac{t}{1620}} \), then \( \left( \frac{1}{2} \right)^{\frac{t}{405}} = x^{4} \):
\[ \frac{1}{5} x^4 + \frac{4}{5} x = \frac{1}{2} \]
Multiplying through by 5:
\[ x^4 + 4x - 2.5 = 0 \]

5. **Solving the Polynomial**:
Using the provided solution \( x = 0.594 \):
\[ \left( \frac{1}{2} \right)^{\frac{t}{1620}} = 0.594 \]

6. **Calculating Time \( t \)**:
Taking logarithm base 10:
\[ \frac{t}{1620} \log_{10} \left( \frac{1}{2} \right) = \log_{10}(0.594) \]
Using \( \log_{10} \left( \frac{1}{2} \right) = -0.3010 \):
\[ \frac{t}{1620} (-0.3010) = -0.5276 \quad (\text{using } \log_{10}(0.594) \approx -0.5276) \]
Thus:
\[ t = 1620 \times \frac{-0.5276}{-0.3010} \approx 1215 \, s \]

### Final Answers:
- The probabilities at \( t = 1620 \, s \) are:
- Probability of alpha particles: \( P_{\alpha} = \frac{1}{9} \)
- Probability of beta particles: \( P_{\beta} = \frac{8}{9} \)

- The time \( t \) when the total number of undecayed nuclei becomes \( \frac{N_0}{2} \) is approximately \( 1215 \, s \).