Okay, so I need to find the minimum resolution required to observe at least 30% of the decays in the CERN Bubble Chamber. The decay is X⁰ → Y⁺ Z⁻, and the proper lifetime of X⁰ is τ₀ = 8×10⁻¹⁶ seconds. The energy in the Bubble Chamber is 27 GeV, and the mass of X⁰ is 3.41 GeV. The choices are A through D, with various lengths given. Hmm, let's think through this step by step.

First, I remember that in particle physics experiments, the resolution of the detector (which determines how precisely we can measure the decay products) affects the number of events we can observe. The higher the resolution, the more precisely we can measure the decay products, and the more decays we can observe. But here, we need the minimum resolution to observe at least 30% of the decays. So, we need to calculate the spread in the decay times due to the finite lifetime and the detector's resolution. 

The key concept here is time dilation and the relativistic effects on the lifetime. The proper lifetime τ₀ is the lifetime in the rest frame of the particle. But in the lab frame (or the frame where the particle is moving), the lifetime τ is longer due to time dilation. The relationship between them is τ = γ τ₀, where γ is the Lorentz factor. 

The Lorentz factor γ is given by γ = E / m, where E is the energy of the particle and m is its rest mass. Wait, no, actually, the total energy E is related to the rest mass m and the kinetic energy. But the formula for γ is E = γ m c², so γ = E / (m c²). But since we're given E and m in GeV, and c² in natural units (where c=1, so units are GeV/c² for mass and GeV for energy), so E (in GeV) = γ m (in GeV). Therefore, γ = E / m = 27 GeV / 3.41 GeV ≈ 7.916. Let me calculate that: 27 divided by 3.41. Let's see, 3.41 times 7 is 23.87, and 27 - 23.87 is 3.13, so 3.13 /3.41 ≈ 0.917. So γ ≈7.917.

So the lifetime in the lab frame is τ = γ τ₀ ≈7.917 *8×10⁻¹⁶ s ≈6.333×10⁻¹⁵ seconds.

But wait, in the lab frame, the particle is moving, so the lifetime is longer. The decay products will have some spatial spread due to the uncertainty in the decay time. The spread in position (Δx) is related to the uncertainty in time (Δt) by the speed of the particle. Since the particle is moving at relativistic speeds, we can approximate its speed as close to c. So Δx ≈ v Δt ≈ c Δt. But Δt here is the uncertainty in the lifetime, which is related to the lifetime itself. 

The spread in the decay positions is due to the decay time uncertainty. The uncertainty principle tells us that the width of the decay (the spread in position) is given by Δx = c τ (1 - 1/(γ²)), but wait, actually, the formula for the decay length is c τ (1 - 1/(γ²)). Wait, no, that's not quite right. Let me think again. The mean decay length is L = v τ = c β γ τ₀. But the uncertainty in the decay time leads to a spread in the positions of the decay products. So the uncertainty in time is the lifetime τ itself. So the spread in position is Δx = c τ. But wait, that would be the case if the decay time is τ, but considering the relativistic effects, maybe it's more accurate to use the lab frame lifetime. So the mean decay length is L = v τ = β c τ. But for the uncertainty in position, it's not the mean, but the spread due to the spread in decay times. The decay time spread is τ, so the corresponding spread in position is Δx = c τ. But since τ is the lab frame lifetime, which is γ τ₀, then Δx = c γ τ₀.

Wait, but the particle is moving at speed v, which is β c. So the mean decay length is β c γ τ₀. But the spread in decay positions would be the uncertainty in position, which is due to the spread in decay times. The spread in decay time is τ, so the spread in position is c times the spread in time. So Δx ≈ c τ = c γ τ₀.

But wait, the question is about the minimum resolution needed to observe at least 30% of the decays. So if the spread in decay positions is Δx, then to observe a certain fraction of decays, the detector's resolution must be better than a certain value. So the resolution (Δx) must be such that the detector can distinguish events within that spread. Wait, no. The resolution is the minimum separation between events that the detector can resolve. So if the spread in decay positions is Δx, then the detector's resolution must be less than or equal to Δx for the detector to be able to distinguish the decays. But the problem is asking for the minimum resolution needed to observe at least 30% of the decays. Hmm, maybe I need to think in terms of the decay width and the detector's ability to resolve events.

Alternatively, perhaps the question is referring to the time resolution of the detector. If the detector can't resolve the decay time with sufficient precision, then it can't detect the decay. But in this case, maybe the spread in the decay time due to the finite lifetime would lead to a spread in the decay positions, and the detector's resolution must be better than that spread to observe all the decays. But the question says "minimum resolution is needed to observe at least 30% of the decays." So perhaps it's about the probability of a decay being observed within a certain spread. If the spread in decay positions is Δx, then the probability of a decay occurring within a detector of size Δx is related to the decay length distribution. 

Wait, maybe it's a Poisson distribution. The number of decays observed would follow a Poisson distribution with mean λ = N × f, where N is the total number of particles and f is the fraction of decays observed. To observe at least 30% of the decays, we need f ≥ 0.3. The probability of observing at least 30% would depend on the spread of the decay positions relative to the detector's resolution. So if the spread is Δx, and the detector has a resolution of δ, then the fraction observed is 1 - e^{-Δx^2/(2δ²)}, assuming a Gaussian distribution. Wait, but the spread might be more like a uniform distribution or something else. Alternatively, perhaps the problem is simpler: the spread in decay positions is proportional to the lifetime and the speed, and the resolution must be less than that spread to observe all decays. But the question is to observe at least 30%, so maybe we need the spread to be such that 30% of the events fall within the detector's resolution. 

Alternatively, perhaps the problem is using the concept of the decay length and the time resolution. Let me think again. The particle's lifetime is τ₀ in its rest frame. In the lab frame, due to time dilation, the lifetime is τ = γ τ₀. The particle is moving at velocity v, so the decay length is L = v τ = β c τ = β c γ τ₀. The spread in the decay positions is the uncertainty in L due to the uncertainty in τ. The uncertainty in τ is τ itself, so the uncertainty in L is ΔL = c γ τ₀. Wait, but that might not be accurate. The uncertainty in the decay time is τ, so the spread in the position is Δx = v Δt = β c τ. Since τ = γ τ₀, then Δx = β c γ τ₀. But β = v/c, and γ = 1/√(1 - β²). So β = sqrt(1 - 1/γ²). But perhaps a better approach is to calculate the spread in the decay positions.

Wait, maybe the key formula here is that the spread in the decay positions (the decay length's uncertainty) is given by Δx = c τ γ. Wait, but τ is the lab frame lifetime, which is γ τ₀. So Δx = c γ τ₀. But I need to verify this. Let's think about it. The decay time in the lab frame is τ = γ τ₀. The velocity of the particle is v = β c. The decay length is L = v τ = β c γ τ₀. But the spread in decay positions (or the uncertainty in L) is due to the uncertainty in τ. Since τ is the lifetime, the uncertainty in L is ΔL = v Δτ. But Δτ is the spread in the lifetime, which is τ itself, because the uncertainty principle tells us that the width of the decay (the spread in time) is related to the lifetime. So Δτ ≈ τ. Therefore, ΔL = v τ = β c τ = β c γ τ₀. 

But the problem is to find the minimum resolution needed to observe at least 30% of the decays. So the detector's resolution (Δx) must be such that the spread in decay positions (ΔL) is less than or equal to the detector's resolution. Wait, but the question is phrased as "minimum resolution needed to observe at least 30% of the decays". Hmm. If the spread in decay positions is ΔL, then the probability of a decay occurring within a detector of size Δx is the fraction of the decay length distribution that lies within Δx. If the spread is larger than the detector's resolution, then fewer events will be observed. To observe at least 30% of the decays, the detector's resolution must be such that the integral of the decay length distribution from 0 to Δx is at least 30%. 

Alternatively, perhaps the problem is simpler. The decay occurs over a certain length, and the detector must resolve this length with a precision such that 30% of the decays are observed. Wait, but how does the resolution relate to the number of observed decays? If the detector has a certain resolution δ, then the number of observed decays is proportional to the number of decay products whose decay positions are within δ of the origin. If the decay positions are spread out over a length Δx, then the fraction observed would be δ/Δx. But this assumes that the decays are uniformly distributed along the path, which might not be the case. Wait, but in reality, the decay positions are spread according to the lifetime distribution. The decay length distribution is exponential in the lab frame, but perhaps the problem is approximating it as a Gaussian. Alternatively, maybe the problem is considering the mean decay length and the standard deviation. 

Alternatively, perhaps the problem is using the concept of the "time resolution" of the detector. The time resolution would determine how precisely the decay time can be measured, which in turn affects the precision of the decay position measurement. The spread in decay positions is related to the uncertainty in the decay time. The uncertainty in decay time Δt is related to the width of the lifetime distribution. For an exponential distribution (which is the case for radioactive decay), the standard deviation of the decay time is equal to the mean lifetime τ. So Δt = τ. The uncertainty in position is then Δx = v Δt = v τ = β c τ. But τ is the lab frame lifetime, which is γ τ₀. So Δx = β c γ τ₀. 

Wait, but β = v/c, and γ = 1/√(1 - β²). So β = sqrt(1 - 1/γ²). Let's compute γ first. As calculated earlier, γ = E/m = 27 GeV / 3.41 GeV ≈7.917. Then 1/γ² ≈1/(62.68)≈0.01595. So 1 - 1/γ² ≈0.98405. Therefore, β = sqrt(0.98405) ≈0.992. So β ≈0.992. Therefore, Δx ≈0.992 * c * γ τ₀. 

Now, τ₀ is the proper lifetime, which is 8×10⁻¹⁶ seconds. γ is ≈7.917. So Δx ≈0.992 * (3×10⁸ m/s) *7.917 *8×10⁻¹⁶ s. Let's compute this step by step. 

First, compute 3×10⁸ m/s * 8×10⁻¹⁶ s = 2.4×10⁻⁸ m. Then multiply by 7.917: 2.4×10⁻⁸ m *7.917 ≈1.899×10⁻⁷ m. Then multiply by 0.992: ≈1.88×10⁻⁷ m. So Δx ≈1.88×10⁻⁷ meters. That's 1.88e-7 m. Let's check the options. The options are A: 2.08e-1 m, B: 2.08e-6 m, C:2.08e-9 m, D:2.08e-3 m. Wait, none of these are 1.88e-7 m. Hmm, but maybe I made a mistake in the calculation.

Wait, let's recompute the numbers carefully. 

First, γ = E/m = 27 / 3.41 ≈7.917. 

Then, 1/γ² = 1/(7.917)^2 ≈1/62.68 ≈0.01595. 

1 - 1/γ² ≈0.98405. 

sqrt(0.98405) ≈0.992. So β ≈0.992. 

Then, Δx = β c γ τ₀. 

c =3×10^8 m/s. 

τ₀ =8×10^-16 s. 

So β c =0.992 *3e8 =2.976e8 m/s. 

Then, β c γ =2.976e8 m/s *7.917 ≈2.36e9 m/s. 

Then, β c γ τ₀ =2.36e9 m/s *8e-16 s =1.888e-6 m. 

Ah, okay, so 1.888e-6 meters, which is approximately 1.89e-6 m. Looking at the options, B is 2.08e-6 m. Close, but not exact. Maybe there's a rounding difference. Let's check the exact calculation with more precise numbers. 

Let me recalculate γ exactly. 

E =27 GeV, m =3.41 GeV. 

γ = E/m =27 /3.41 = let's compute this precisely. 

3.41 *7 =23.87, 27 -23.87=3.13. 

3.13 /3.41 =0.9177. 

So γ =7.9177. 

Then, 1/γ² =1/(7.9177)^2. 

7.9177^2 = (7 +0.9177)^2 =49 + 2*7*0.9177 +0.9177². 

7*0.9177=6.4239, so 2*7*0.9177=12.8478. 

0.9177²≈0.842. 

So total 49 +12.8478 +0.842≈62.69. 

So 1/62.69≈0.015945. 

So 1 -1/γ²≈0.984055. 

sqrt(0.984055)= let's compute this. 

sqrt(0.984055). Let's see, 0.992^2=0.984064. So sqrt(0.984055)=approximately 0.992. 

So β≈0.992. 

Now, compute β c γ τ₀: 

β c =0.992 *3e8 =2.976e8 m/s. 

γ =7.9177. 

So β c γ =2.976e8 *7.9177 ≈2.976e8 *7.9177. 

Let's compute 2.976 *7.9177. 

2.976 *7 =20.832. 

2.976 *0.9177≈2.976*0.9=2.6784; 2.976*0.0177≈0.0526. 

Total ≈2.6784 +0.0526≈2.731. 

So total 20.832 +2.731≈23.563. 

So β c γ ≈23.563e8 m/s. 

Then multiply by τ₀=8e-16 s: 

23.563e8 m/s *8e-16 s = (23.563 *8) *1e-8 m =188.504 *1e-8 m =1.88504e-6 m. 

So approximately 1.885e-6 meters. 

The options are:

A. 2.08*1e-1 m → 0.208 m

B. 2.08*1e-6 m →0.00000208 m

C.2.08*1e-9 m →0.00000000208 m

D.2.08*1e-3 m →0.00208 m

So 1.885e-6 m is approximately 1.885 x10^-6 m. Option B is 2.08e-6 m. The calculated value is close to B but a bit lower. But perhaps there's a mistake in the approach. Wait, maybe the problem is using a different formula. Let me think again.

Another approach: The uncertainty in the decay time is τ, so the uncertainty in position is Δx = v τ. But τ is the lab frame lifetime, which is γ τ₀. So Δx = v γ τ₀. But v = β c. So Δx = β c γ τ₀. Which is the same as before. So the calculation seems correct. 

But the options don't have 1.88e-6 m. The closest is B:2.08e-6 m. Maybe the calculation is slightly off due to more precise values. Let's check with exact numbers. 

Compute γ =27 /3.41 =7.917 approximately. 

Compute β = sqrt(1 -1/γ²). 

1/γ² =1/(7.917)^2. 

7.917^2 =62.68. 

So 1/62.68≈0.015945. 

1 -0.015945=0.984055. 

sqrt(0.984055)=0.992. 

So β=0.992. 

Then β c γ τ₀ =0.992 *3e8 m/s *7.917 *8e-16 s. 

Compute 0.992 *7.917=7.917*0.992. 

7*0.992=6.944. 0.917*0.992=0.909. 

Total≈6.944 +0.909=7.853. 

Then 7.853 *3e8=2.3559e9 m/s. 

Multiply by 8e-16 s: 2.3559e9 *8e-16 =1.8847e-6 m. 

So exactly 1.8847e-6 m. So approximately 1.885e-6 m. The options don't have this exact value, but option B is 2.08e-6 m. There's a discrepancy here. Maybe the problem is using a different formula. Alternatively, perhaps the spread in decay positions is different. 

Wait, maybe the decay length is L = γ v τ₀. Because in the rest frame, the decay length would be L = v τ₀, but in the lab frame, it's multiplied by gamma. Wait, no, in the lab frame, the lifetime is gamma tau0, so L = v * gamma tau0. Which is the same as before. So the calculation seems correct. 

Alternatively, perhaps the problem is considering the mean path length, but that's not relevant here. 

Alternatively, maybe the problem is using the concept of the detector's resolution being the same as the spread in decay positions. So if the detector can't resolve positions better than the spread, then the number of observed decays would be related to the integral of the decay distribution. But perhaps the problem simplifies it by assuming that the detector's resolution must be equal to the spread in decay positions. But in reality, it's more complicated because the decay positions are distributed according to the lifetime, so the fraction observed depends on the detector's resolution. 

If we assume that the detector's resolution must be equal to the spread in decay positions (Δx), then the minimum resolution would be the spread, which is approximately 1.88e-6 m. But since the options don't have this exact value, the closest one is B:2.08e-6 m. However, there's a difference of about 0.2e-6 m. Maybe there's a miscalculation. Let me check the calculation again. 

Wait, perhaps the original calculation used a different gamma. Let's compute gamma more accurately. 

E =27 GeV. m=3.41 GeV. 

gamma = E/m =27 /3.41. Let's compute this precisely. 

3.41 ×7 =23.87. 27 -23.87=3.13. 

3.13 /3.41 =0.9177. 

So gamma =7.9177. 

Then 1/gamma² =1/(7.9177)^2. 

Compute 7.9177^2:

7 *7 =49. 

7 *0.9177 =6.4239. 

0.9177 *7 =6.4239. 

0.9177 *0.9177 ≈0.842. 

So (7 +0.9177)^2 =49 + 2*7*0.9177 +0.9177² =49 +12.8478 +0.842 ≈62.69. 

So 1/62.69 ≈0.015945. 

1 -1/62.69 ≈0.984055. 

sqrt(0.984055) ≈0.992. 

So beta ≈0.992. 

Then, beta * c * gamma * tau0 =0.992 *3e8 *7.9177 *8e-16. 

Compute 0.992 *7.9177 ≈7.853. 

7.853 *3e8 =2.3559e9. 

2.3559e9 *8e-16 =1.8847e-6 m. 

So the calculation is correct. 

But the options don't have this. The closest option is B:2.08e-6 m. Maybe there's a miscalculation in the problem statement. Alternatively, perhaps the problem uses the formula Δx = c * τ₀ * γ * β. Which is the same as before. 

Alternatively, maybe the problem is considering the mean decay length. The mean decay length is L = v * τ = β c * γ τ₀. Which is the same as before. So the spread is the same. 

Alternatively, perhaps the problem is using a different approach. Let's think about the decay length distribution. The decay length follows an exponential distribution. The probability density function is f(x) = (1/τ) e^{-x/τ}, where τ is the mean lifetime. The probability that a decay occurs within a distance x is 1 - e^{-x/τ}. To have at least 30% of the decays observed, we need 1 - e^{-x/Δx} ≥0.3. Solving for x, where Δx is the detector resolution. 

So 1 - e^{-x/Δx} =0.3 → e^{-x/Δx}=0.7 → -x/Δx = ln(0.7) → x = -Δx ln(0.7). 

But we need the detector to observe at least 30% of the decays, so the fraction is 0.3. Therefore, the detector resolution must be at least Δx = -x / ln(0.7). 

Wait, but if the spread in decay positions is Δx, then to observe at least 30% of the decays, the detector's resolution must be better than the spread. Wait, no. The spread in decay positions is the standard deviation (or mean) of the decay lengths. The fraction observed would depend on the number of events within the detector's acceptance. If the detector can only accept events within a certain range, then the fraction observed depends on the detector's resolution. 

But perhaps the problem is simplified by assuming that the detector's resolution must be equal to the spread in decay positions. But according to the calculation, the spread is 1.88e-6 m, and the closest option is B:2.08e-6 m. 

Alternatively, maybe the problem is considering the half-life instead of the proper lifetime. Wait, the problem states "proper lifetime of X⁰ is 8×10⁻¹⁶ s". So that's τ₀. The lifetime in the lab frame is γ τ₀. 

Wait, maybe the problem is using the decay length formula as c * tau0 * gamma * beta. Which is the same as before. 

Alternatively, perhaps the problem is using the formula for the uncertainty in position due to the finite lifetime, which is Δx = c * gamma * tau0. But without the beta, because maybe it's assuming the particle is moving at speed c, which isn't possible, but sometimes approximated. 

If that's the case, Δx = c * gamma * tau0. Which would be 3e8 m/s *7.917 *8e-16 s. 

3e8 *7.917 ≈2.3751e9. 

2.3751e9 *8e-16 =1.900e-6 m. 

That's approximately 1.9e-6 m, which is close to option B (2.08e-6 m). 

But even this approach ignores the beta factor, which is part of the formula. However, in reality, the velocity is v = beta c, so the spread should include beta. But maybe in the problem, they are simplifying by ignoring beta. 

Alternatively, perhaps the problem is considering the mean path length, which is L = c tau gamma. Which is similar to what we have. 

In any case, the calculated value is approximately 1.88e-6 m, which is closest to option B:2.08e-6 m. However, the discrepancy could be due to rounding in the calculation. Let's check with more precise values. 

gamma =27 /3.41 = exactly 27 divided by 3.41. Let's compute this exactly. 

3.41 ×7 =23.87. 

3.41 ×7.9177 =27. 

So gamma is exactly 27/3.41. Let's compute it precisely. 

3.41 ×7 =23.87. 

3.41 ×0.9177 = let's compute 3.41 ×0.9=3.069, 3.41 ×0.0177=0.0603. So total 3.069 +0.0603=3.1293. 

So 3.41 ×7.9177=23.87 +3.1293=27. So gamma is exactly 7.9177. 

Then 1/gamma² =1/(7.9177)^2. Let's compute 7.9177 squared. 

7.9177 ×7.9177. Let's compute this step by step. 

7 ×7 =49. 

7 ×0.9177 =6.4239. 

0.9177 ×7 =6.4239. 

0.9177 ×0.9177 ≈0.842. 

So total is 49 +6.4239 +6.4239 +0.842 ≈49 +12.8478 +0.842 ≈62.6898. 

So 1/gamma² ≈1/62.6898 ≈0.015945. 

1 -1/gamma² ≈0.984055. 

sqrt(0.984055) ≈0.992. 

So beta ≈0.992. 

Then beta * c * gamma * tau0 =0.992 *3e8 *7.9177 *8e-16. 

Compute 0.992 *7.9177 ≈7.853. 

7.853 *3e8 =2.3559e9. 

2.3559e9 *8e-16 =1.8847e-6 m. 

So the exact calculation gives 1.8847e-6 m. The options don't have this. The closest is B:2.08e-6 m. But perhaps the problem expects us to ignore the beta factor and use gamma * c * tau0. 

If we compute gamma * c * tau0:7.9177 *3e8 m/s *8e-16 s. 

7.9177 *3e8 =2.3753e9. 

2.3753e9 *8e-16 =1.9002e-6 m. 

Which is approximately 1.9e-6 m. Still not matching exactly, but closer. 

Alternatively, maybe the problem is using the mean path length formula without considering the beta factor. 

Alternatively, perhaps the problem is considering the mean path length as c * tau0 * gamma. Which would be 3e8 *8e-16 *7.9177 =1.9002e-6 m. 

Again, leading to approximately 1.9e-6 m. 

But the options don't have this. The closest is B:2.08e-6 m. 

Alternatively, perhaps the problem is considering the uncertainty in the decay time as the spread in positions. So delta x = c * tau * beta gamma. Which is the same as before. 

Alternatively, maybe there's a miscalculation in the problem. Let's check the given options again. 

The options are:

A. 2.08*1e-1 m →0.00208 m

B. 2.08*1e-6 m →0.00000208 m

C.2.08*1e-9 m →0.00000000208 m

D.2.08*1e-3 m →0.00208 m

Wait, the calculated value is approximately 1.88e-6 m, which is 0.00000188 m, which is 1.88e-6 m. The closest option is B:2.08e-6 m, which is 2.08e-6 m. The difference is about 0.2e-6 m. But perhaps the problem expects us to round to the nearest given option. Alternatively, perhaps there's a miscalculation in the problem. 

Alternatively, maybe the problem is considering the uncertainty in the decay time as tau, and the spread in position is c * tau. But tau in the lab frame is gamma tau0. So spread is c * gamma tau0. Which is the same as before. 

Alternatively, perhaps the problem is considering the half-life instead of the proper lifetime. Wait, the problem states "proper lifetime of X⁰ is 8×10⁻¹⁶ s". So tau0 is 8e-16 s. 

Alternatively, perhaps the problem is using the formula for the mean decay length, which is L = gamma c tau0. Which is same as before. 

Wait, but in reality, the decay length distribution is exponential. The probability that a decay occurs at a distance x is (1/L) e^{-x/L}, where L is the mean decay length. The cumulative distribution function is 1 - e^{-x/L}. To have at least 30% of the decays within the detector's resolution, we need 1 - e^{-Δx/Δx} ≥0.3. Wait, but Δx here is the detector's resolution. Wait, no. If the detector's resolution is Δx, then the fraction observed would be 1 - e^{-Δx/Δx} =1 - e^{-1}=1 -1/e≈0.632. Which is 63.2% of the decays. That's more than 30%. So that can't be the case. 

Alternatively, perhaps the fraction observed is given by the ratio of the detector's resolution to the spread. For example, if the spread is sigma, then the fraction within ±k sigma is given by the error function. But the problem states "minimum resolution is needed to observe at least 30% of the decays". So if the spread is sigma, then the fraction observed would be the integral from -infty to +infty of the decay length distribution. Wait, that doesn't make sense. 

Alternatively, perhaps the problem is considering the time resolution of the detector. The time resolution must be such that the number of decay events observed is at least 30% of the total. The time resolution delta_t must be such that the number of decays with |t| <= delta_t is at least 30% of the total. 

The number of decays observed is proportional to the integral of the decay length distribution over the detector's resolution. The decay length distribution is exponential: f(x) = (1/L) e^{-|x|/L}, where L = gamma c tau0. 

The fraction of decays observed within a detector resolution delta_x is 1 - e^{-delta_x/L}. To have at least 30% observed, we need 1 - e^{-delta_x/L} ≥0.3. Solving for delta_x:

1 - e^{-delta_x/L} ≥0.3 → e^{-delta_x/L} ≤0.7 → -delta_x/L ≥ ln(0.7) → delta_x ≤ -L ln(0.7). 

But since delta_x must be positive, we have delta_x ≤ -L ln(0.7). 

Compute ln(0.7)= -0.35667. So delta_x ≤ -L (-0.35667) =0.35667 L. 

So delta_x ≤0.35667 L. 

But L = gamma c tau0. So delta_x ≤0.35667 * gamma c tau0. 

Compute delta_x ≤0.35667 *7.9177 *3e8 *8e-16. 

First compute 0.35667 *7.9177 ≈2.833. 

Then 2.833 *3e8 ≈8.499e8. 

Then 8.499e8 *8e-16 =6.799e-7 m. 

So delta_x ≤6.799e-7 m. 

Wait, that's approximately 6.8e-7 m. The closest option is B:2.08e-6 m. But 6.8e-7 is 0.68e-6 m, which is not close to any option. So this approach doesn't seem to fit. 

Alternatively, perhaps the problem is considering the time resolution delta_t, and the corresponding position resolution delta_x = delta_t * v. 

If the detector's time resolution is delta_t, then the corresponding position resolution is delta_x = delta_t * v. 

The number of decays observed within delta_t is the integral of the time distribution from t=0 to t=delta_t. The time distribution is exponential: f(t) = (1/tau) e^{-t/tau}, where tau is the mean lifetime. 

The number of decays observed within delta_t is the integral from 0 to delta_t of f(t) dt =1 - e^{-delta_t/tau}. To have at least 30% observed, 1 - e^{-delta_t/tau} ≥0.3. Solving for delta_t:

1 - e^{-delta_t/tau} ≥0.3 → e^{-delta_t/tau} ≤0.7 → -delta_t/tau ≥ ln(0.7) → delta_t ≤ -tau ln(0.7). 

So delta_t ≤ -tau ln(0.7). 

Compute ln(0.7)= -0.35667. So delta_t ≤0.35667 tau. 

tau is the mean lifetime in the lab frame, which is gamma tau0. 

So delta_t ≤0.35667 * gamma tau0. 

Then delta_x = delta_t * v =0.35667 * gamma tau0 * beta c. 

But beta c =v. So delta_x =0.35667 * gamma tau0 * beta c. 

Which is the same as before. 

Compute delta_x =0.35667 *7.9177 *8e-16 *0.992 *3e8. 

First compute 0.35667 *7.9177 ≈2.833. 

Then 2.833 *8e-16 ≈2.266e-15. 

2.266e-15 *0.992 ≈2.243e-15. 

2.243e-15 *3e8 ≈6.729e-7 m. 

Again, around 6.7e-7 m, which is 0.67e-6 m. Still not matching the options. 

So this approach also leads to a different value. 

But the problem states "minimum resolution is needed to observe at least 30% of the decays". It might be that the problem is considering the spread in decay positions as the detector's resolution. Therefore, the detector's resolution must be equal to the spread in decay positions. 

From the earlier calculation, the spread is approximately 1.88e-6 m. The closest option is B:2.08e-6 m. 

But why is there a discrepancy? Maybe the problem uses different values for gamma or beta. Let's check the calculation again. 

Wait, perhaps the problem is using the formula for the uncertainty in position due to the uncertainty in the decay time. The formula is Δx = c Δt. 

Where Δt is the uncertainty in the decay time. For a Gaussian distribution, the standard deviation of the decay time is equal to the mean lifetime τ. But for a Poisson process, the decay time uncertainty is τ. However, in reality, the decay time uncertainty is τ. 

But the spread in decay positions would then be Δx = c Δt = c τ. But in the lab frame, Δt = γ τ₀. So Δx = c γ τ₀. 

Wait, this would be a simpler approach. If the problem considers the uncertainty in decay time as τ (lab frame), then Δx = c γ τ₀. 

Compute this: c=3e8 m/s, γ=7.9177, τ₀=8e-16 s. 

Δx =3e8 *7.9177 *8e-16 =3e8 *7.9177=2.375e9. 2.375e9 *8e-16=1.9e-6 m. 

Which is approximately 1.9e-6 m. Again, close to option B (2.08e-6 m). 

So perhaps the problem is simplifying by considering the uncertainty in the decay time as τ (lab frame), and then the spread in position is c τ. But since tau in the lab frame is gamma tau0, then Δx = c gamma tau0. Which is the same as before. 

But the correct calculation based on the uncertainty principle or detection resolution would involve the lifetime spread. However, the options given suggest that the correct answer is B:2.08e-6 m. But according to the calculations, the spread is approximately 1.88e-6 m. 

Alternatively, perhaps the problem is using the formula for the half-life instead of the proper lifetime. Let's check. The half-life T_1/2 = tau0 ln(2). So T_1/2 =8e-16 *0.693≈5.545e-16 s. 

But the gamma factor would then be gamma =E/m=27/3.41≈7.917. So T_1/2 lab frame=gamma * T_1/2 proper≈7.917*5.545e-16≈4.39e-15 s. 

But I don't see how this relates to the problem. 

Alternatively, perhaps the problem is considering the detector's resolution as the spread in decay lengths, which is gamma c tau0. Which is approximately 1.9e-6 m. 

Given that the closest option to 1.9e-6 m is B:2.08e-6 m, I think the intended answer is B. 

But why is there a discrepancy? Maybe there's a miscalculation in the problem. Alternatively, perhaps the problem uses a different formula. 

Wait, another approach: The decay length is L = gamma beta c tau0. The standard deviation of the decay length is L. To observe at least 30% of the decays, the detector's resolution must be at least equal to the standard deviation. So the minimum resolution is L = gamma beta c tau0. Which is the same as before. 

Compute gamma beta c tau0:7.9177 *0.992 *3e8 *8e-16. 

7.9177 *0.992 ≈7.853. 

7.853 *3e8 ≈2.3559e9. 

2.3559e9 *8e-16 =1.8847e-6 m. 

Again, same result. 

Therefore, the correct answer should be approximately 1.88e-6 m, which is closest to option B:2.08e-6 m. 

But given the options, perhaps the intended answer is B. Therefore, the answer is B.
</think>

To determine the minimum resolution required to observe at least 30% of the decays in the given scenario, we analyze the problem using relativistic time dilation and the spread in decay positions. The key steps are as follows:

1. **Proper Lifetime and Lorentz Factor**:  
   The proper lifetime of the particle \( X^0 \) is given as \( \tau_0 = 8 \times 10^{-16} \, \text{s} \). The energy of the particle in the lab frame is \( E = 27 \, \text{GeV} \), and the mass of \( X^0 \) is \( m = 3.41 \, \text{GeV}/c^2 \). The Lorentz factor \( \gamma \) is calculated as:  
   \[
   \gamma = \frac{E}{m} = \frac{27}{3.41} \approx 7.917
   \]  
   The lab frame lifetime is then:  
   \[
   \tau = \gamma \tau_0 = 7.917 \times 8 \times 10^{-16} \, \text{s} \approx 6.334 \times 10^{-15} \, \text{s}.
   \]

2. **Spread in Decay Positions**:  
   The spread in decay positions is determined by the uncertainty in the decay time, which is related to the lab frame lifetime \( \tau \). The spread in decay positions is given by:  
   \[
   \Delta x = \beta c \tau
   \]  
   where \( \beta = v/c \) is the velocity of the particle divided by the speed of light. Using \( \gamma = \frac{1}{\sqrt{1 - \beta^2}} \), we find:  
   \[
   \beta = \sqrt{1 - \frac{1}{\gamma^2}} \approx \sqrt{1 - \frac{1}{7.917^2}} \approx 0.992
   \]  
   Substituting \( \beta \approx 0.992 \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \tau \approx 6.334 \times 10^{-15} \, \text{s} \), the spread in decay positions is:  
   \[
   \Delta x = 0.992 \times 3 \times 10^8 \times 6.334 \times 10^{-15} \approx 1.885 \times 10^{-6} \, \text{m}.
   \]

3. **Minimum Resolution**:  
   The minimum resolution required to observe at least 30% of the decays corresponds to the spread in decay positions \( \Delta x \). Since the problem asks for the minimum resolution needed, the detector's resolution must be at least equal to the spread in decay positions. The calculated spread in decay positions is approximately \( 1.885 \times 10^{-6} \, \text{m} \). Among the given options, the closest value is:  
   \[
   \text{B. } 2.08 \times 10^{-6} \, \text{m}.
   \]

Thus, the minimum resolution required to observe at least 30% of the decays is **B: \( 2.08 \times 10^{-6} \, \text{m} \)**.

ANSWER: B