Calculation:

1. After 10 years (1 half-life), the mass remaining is 400g / 2 = 200g.
2. After another 10 years (2 half-lives), the mass remaining is 200g / 2 = 100g.
3. After a further 10 years (3 half-lives), the mass remaining is 100g / 2 = 50g.

Answer: After 30 years, 50g of the isotope will remain.

Solution:

1. The relationship between activity and time is given by: A(t) = A₀e^-kt, where A₀ is the initial activity, k is the decay constant, and t is time.
2. We are given that A(200) = A₀/3 and t = 200 s. Substituting these values: A₀/3 = A₀e^-200k
3. Divide both sides by A₀: 1/3 = e^-200k
4. Take the natural logarithm of both sides: ln(1/3) = -200k => -1.099 = -200k
5. Solve for k: k = -1.099 / -200 = 0.005495 s^-1
6. The half-life (T_1/2) is given by: T_1/2 = ln(2) / k
7. Substitute the value of k: T_1/2 = 0.693 / 0.005495 = 126 s
8. Therefore, the half-life of the radioactive substance is 126 seconds.

Solution:

1. The number of atoms remaining after time t is given by the equation: N(t) = N₀e^-kt, where N₀ is the initial number of atoms, k is the decay constant, and t is time.
2. We are given N₀ = 1.6 x 10⁷ and N(100) = 4.0 x 10⁶, and t = 100 s. Substituting these values into the equation: 4.0 x 10⁶ = 1.6 x 10⁷e^-100k
3. Divide both sides by 1.6 x 10⁷: (4.0 x 10⁶) / (1.6 x 10⁷) = e^-100k => 0.25 = e^-100k
4. Take the natural logarithm of both sides: ln(0.25) = -100k => -1.386 = -100k
5. Solve for k: k = -1.386 / -100 = 0.01386 s^-1
6. The half-life (T_1/2) is given by the equation: T_1/2 = ln(2) / k
7. Substitute the value of k: T_1/2 = 0.693 / 0.01386 = 49.5 s
8. Therefore, the half-life of isotope X is approximately 49.5 seconds.