Okay, so I need to figure out the comoving distance of a quasar based on the given parameters from the Lambda-CDM model. Let me start by recalling what I know about cosmology and the Lambda-CDM model. 

The question states that observations in the near infrared and optical spectrum of a quasar show a peak at 790 nm, and the flux drops significantly at shorter wavelengths. I'm not entirely sure how that relates to the distance, but maybe it's about the light curve or something about the expansion of the universe affecting the observed wavelength? Wait, maybe that's a redshift consideration. If the flux drops at shorter wavelengths, that might imply that the object is moving away, so redshifted. But the quasar is emitting light, and we observe a peak at 790 nm. But quasars typically emit in the UV, so maybe the peak here is in the optical or near-infrared. Hmm, but I'm not sure if that's directly relevant to calculating the distance. Maybe the redshift is inferred from the observed wavelength compared to the rest frame. Let me think. 

Wait, the peak wavelength is 790 nm. But quasars usually have emission lines in the UV, but their overall spectrum extends into the optical and near-infrared. But if the peak is at 790 nm (which is 0.79 microns), that's in the near-infrared. But how does that help us find the distance? Maybe the peak wavelength is related to the redshift. Because if the quasar is moving away, the light we observe would be redshifted. So if the rest wavelength (emitted) is, say, in the UV, and we observe it at 790 nm, then we can calculate the redshift z. But the question doesn't give the rest wavelength. Wait, maybe the peak is the observed wavelength, and the rest wavelength is known? Or perhaps the peak is at 790 nm, but the rest wavelength is the same, so the redshift can be calculated. Wait, maybe that's the key here. 

If the quasar's emission peak is at a certain rest wavelength, and we observe it at 790 nm, then the redshift z is (observed wavelength / rest wavelength) - 1. But the problem is that we don't know the rest wavelength. Wait, but quasars are variable sources, and their emission lines can be broad and sometimes have specific features. But maybe in this case, the peak is the maximum of the emission in the observed spectrum, which could indicate the rest wavelength. However, without knowing the rest wavelength, it's impossible to calculate z. Hmm, maybe I'm missing something here. 

Wait, perhaps the flux dropping at shorter wavelengths is a clue. If the flux drops at shorter wavelengths (lower wavelengths, higher frequencies), that could indicate that the object is moving away at a certain redshift, causing the light to be redshifted. But again, without knowing the rest frame wavelength, I can't compute the redshift directly. Maybe the peak wavelength is the rest wavelength, so if we can find the redshift from that, but I don't know the rest wavelength. Alternatively, maybe the peak wavelength is 790 nm, and that's the observed wavelength, so if we assume the rest wavelength is, say, 400 nm (a typical UV emission), then z would be (790/400) -1 ≈ 0.975. But the problem is that the question doesn't specify the rest wavelength. Maybe the question is more about the cosmological distance calculation based on the given cosmological parameters, not the redshift. Wait, but the Hubble constant, matter density, dark energy density, etc., are given. So perhaps the redshift can be determined from the observed wavelength, and then we can compute the comoving distance using the cosmological parameters. But without knowing the rest wavelength, how do we get the redshift? 

Wait, maybe the peak at 790 nm is the rest wavelength of the quasar, and we just need to calculate the comoving distance assuming that redshift. But the problem is that quasars have broad emission lines, and the peak might not be at a specific rest wavelength. Alternatively, maybe the peak is the observed wavelength, and the rest wavelength is known from the quasar's spectral features. But since the rest wavelength isn't given, perhaps the question is more about the cosmological distance formula. Let me check the problem again. 

The question is asking about the comoving distance for a scale factor a=1, which is the present time. So the comoving distance is the distance we would measure if the universe were static, ignoring the expansion. But with the given cosmological parameters (H0=70 km/s/Mpc, Omega_m=0.3, Omega_lambda=0.7, flat universe), we can compute the comoving distance using standard cosmological integrals. 

Wait, but how does the observed wavelength relate to the comoving distance? Because the observed wavelength is affected by the redshift, which is related to the scale factor. The observed wavelength lambda_obs = lambda_rest * (1 + z) = lambda_rest * a_emit / a_obs. Since a_emit is the scale factor at emission, and a_obs is the scale factor now (a=1), so z = (a_emit / a_obs) -1 = (a_emit) -1. But without knowing the rest wavelength, we can't find z. So maybe the peak at 790 nm is the observed wavelength, and we need to use some other method to find the redshift. Alternatively, perhaps the question is implying that the observed peak is due to the expansion of the universe, and we can use the cosmological parameters to compute the comoving distance directly without needing the redshift. But I'm not sure. 

Alternatively, perhaps the question is a trick question, and the answer is based on the cosmological parameters without needing the redshift. Let me think. The comoving distance D_c is given by the integral from 0 to z of c/(H0 * E(z')) dz', where E(z) = sqrt(Omega_m*(1+z)^3 + Omega_lambda*(1+z)^0). But without knowing z, how do we compute D_c? So perhaps the observed peak wavelength is a redshift indicator, and we can compute z from that, then use z to compute D_c. But again, without knowing the rest wavelength, we can't compute z. 

Wait, maybe the peak at 790 nm is the rest wavelength of the quasar. Let's assume that. So if the rest wavelength is 790 nm, and the observed wavelength is 790 nm, then z=0. But that can't be right because the problem states that at shorter wavelengths, the flux drops. So if the rest wavelength is, say, 400 nm (UV), then the observed peak at 790 nm would correspond to z = (790/400) -1 ≈ 0.975. But without knowing the rest wavelength, we can't do that. 

Alternatively, maybe the question is about the luminosity distance, but the problem specifically asks about the comoving distance. So perhaps the answer is based on the cosmological parameters alone. Let me check the options. The options are 6, 7, 8, 9 Gpc. The Hubble constant is 70 km/s/Mpc, which is a common value. The matter density is 0.3, dark energy 0.7. 

In a flat universe with these parameters, the comoving distance can be calculated using the formula. The comoving distance D_c is given by (c / H0) * ∫ (from 0 to z) [1 / E(z')] dz', where E(z) = sqrt(Ω_m*(1+z)^3 + Ω_Λ). 

But without knowing z, we can't compute this. So perhaps the question is implying that the observed wavelength is the rest wavelength, and the redshift is determined from that. Let's try that. If the rest wavelength is, say, 400 nm (a typical quasar emission line), then observed wavelength is 790 nm. So z = 790/400 -1 = 0.975. Then using this z, compute D_c. 

Alternatively, maybe the peak is in the optical, so rest wavelength is longer. Let's say the rest wavelength is 1.2 microns (near-infrared), so observed at 0.79 microns. Then z = 0.79 / 1.2 -1 ≈ 0.658 -1 = negative, which doesn't make sense. So that can't be. Alternatively, maybe the rest wavelength is in the UV, say 0.5 microns (500 nm), observed at 0.79 microns. Then z = 0.79 / 0.5 -1 ≈ 0.58. 

But without knowing the rest wavelength, this is speculative. Maybe the question is trying to trick us, and the answer is based on the cosmological parameters without needing the redshift. Let's think. The Hubble constant H0=70 km/s/Mpc. The comoving distance at high z would be roughly D_c ≈ (c/H0) * (1/(1 - q0 z/2)) ) for small z, but that's an approximation. Wait, but if the quasar is very far, z would be large, but the options are up to 9 Gpc. Let's convert H0 to s^-1. H0 = 70 km/s/Mpc = 70e3 m / (3.086e22 m) s^-1 ≈ 70e3 / 3.086e22 ≈ 2.27e-18 s^-1. Then c/H0 = 3e8 m/s / 2.27e-18 s^-1 ≈ 1.32e26 m ≈ 13.2 Mpc. Wait, that can't be right. Wait, wait, let's do the calculation correctly. H0 is 70 km/s/Mpc. So 1 Mpc = 3.086e22 m. So H0 = 70 km/s/Mpc = 70e3 m / (3.086e22 m) s^-1 = 70e3 / 3.086e22 ≈ 2.27e-18 s^-1. Then c/H0 = 3e8 m/s / 2.27e-18 s^-1 ≈ 1.32e26 m. Convert that to Mpc: 1 Mpc is 3.086e22 m. So 1.32e26 / 3.086e22 ≈ 427 Mpc. So c/H0 ≈ 427 Mpc. 

But that's just the Hubble distance. The comoving distance at high z is roughly 2 * Hubble distance / (1 - sqrt(1 - Omega_m - Omega_lambda)) ), but that's getting complicated. Alternatively, for a flat universe with Omega_m=0.3 and Omega_lambda=0.7, the integral for comoving distance can be approximated. 

Alternatively, I can recall that in a flat universe with these parameters, the comoving distance at high redshifts is about D_c ≈ 1.5 * Hubble_distance * (1 + z) / (1 + sqrt(Omega_m (1+z)^3 + Omega_lambda)) ), but I'm not sure. 

Alternatively, maybe the answer is based on the luminosity distance instead of comoving distance. But the question specifically asks for comoving distance. 

Wait, maybe the question is suggesting that the quasar's observed peak wavelength is due to the expansion, and we can compute the redshift from that. But without knowing the rest wavelength, this is impossible. So perhaps the question is a trick and the answer is based on the cosmological parameters without needing the redshift. But I don't see how. 

Alternatively, perhaps the question is referring to the fact that the flux drops at shorter wavelengths, implying that the object is moving away, hence redshifted, and we can use the observed wavelength to compute the redshift. But again, without the rest wavelength, it's not possible. 

Wait, maybe the peak is a certain rest wavelength that corresponds to a particular redshift. For example, if the peak is at 790 nm, and assuming that's the rest wavelength, then z would be 0, which can't be. Or if the rest wavelength is 400 nm (0.4 microns), then z=0.975. Then comoving distance would be calculated for z=0.975. Let's compute that. 

The comoving distance formula is D_c = (c / H0) * ∫₀^z [1 / sqrt(Omega_m*(1+z')^3 + Omega_lambda)] dz'. 

Given Omega_m=0.3, Omega_lambda=0.7. Let's compute the integral numerically. For z=0.975. 

But doing the integral by hand is tedious. Maybe approximate. Let's use the approximation for D_c in terms of z. For a flat universe with these parameters, the integral can be approximated. 

Alternatively, use the formula for comoving distance. There's a standard approximation for comoving distance in terms of redshift. For example, for z << 1, D_c ≈ cz / H0. But for higher z, it's more complicated. 

Alternatively, use the relation between luminosity distance and comoving distance. Luminosity distance D_L = (1 + z) * D_c. But without knowing D_L, that's not helpful. 

Alternatively, look up an online calculator or use an approximation formula. The integral for D_c can be approximated numerically. Let's try to compute it for z=0.975. 

The integrand is 1 / sqrt(0.3*(1+z')^3 + 0.7). 

We need to integrate this from z'=0 to z'=0.975. 

Let me make a table of approximate values at different z' and compute the integral using Simpson's rule or something. 

Alternatively, for a rough estimate, we can use the following approximation for comoving distance in a flat universe:

D_c = (c / H0) * 2/(sqrt(Omega_lambda)) * arcsinh( sqrt(Omega_m / Omega_lambda) (1 + z) )

But I'm not sure if that's accurate. Let me check. Wait, the general formula for comoving distance in a flat universe (Omega_m + Omega_lambda =1) is:

D_c = (c / H0) * ∫₀^z [1 / sqrt(Omega_m (1 + z')^3 + Omega_lambda)] dz'

This integral can be expressed in terms of hypergeometric functions or elliptic integrals, but maybe there's an approximate formula. 

Alternatively, use the following approximation for the integral:

∫₀^z [1 / sqrt(a (1 + z')^3 + b)] dz' ≈ (2/(3a)) * [ (1 + z)^{3/2} - 1 ] / sqrt(b + a (1 + z)^3 )

But I'm not sure. Alternatively, use substitution. Let u = 1 + z', then du = dz'. Then the integral becomes ∫₁^{1+z} [1 / sqrt(Omega_m u^3 + Omega_lambda)] du. 

Hmm, perhaps using substitution v = u^3. Then dv = 3u^2 du, but that might not help. Alternatively, for small Omega_m, but here Omega_m=0.3 is not that small. 

Alternatively, use a series expansion. For small z, but z=0.975 is not small. 

Alternatively, use the approximation from the book "Cosmology" by Weinberg or similar. Wait, I think the comoving distance for a flat universe can be written as:

D_c = (c / H0) * (2 / sqrt(Omega_lambda)) ) * arcsinh( sqrt(Omega_m/Omega_lambda) (1 + z) )

Wait, let me check. If z is small, then (1+z) ≈ 1, so arcsinh(sqrt(Omega_m/Omega_lambda)) ≈ arcsinh(sqrt(0.3/0.7)) ≈ arcsinh(0.6547) ≈ 0.673. But for larger z, this might not hold. Let's try plugging in z=0.975. 

Compute sqrt(Omega_m/Omega_lambda) = sqrt(0.3/0.7) ≈ sqrt(0.4286) ≈ 0.6547. Then (1 + z) = 1.975. So sqrt(Omega_m/Omega_lambda)*(1 + z) ≈ 0.6547 * 1.975 ≈ 1.293. Then arcsinh(1.293). Let's compute that. 

arcsinh(x) = ln(x + sqrt(x^2 +1)). So ln(1.293 + sqrt(1.293^2 +1)) = ln(1.293 + sqrt(1.672 +1)) = ln(1.293 + sqrt(2.672)) ≈ ln(1.293 + 1.635) ≈ ln(2.928) ≈ 1.074. 

Then D_c ≈ (c / H0) * (2 / sqrt(0.7)) * 1.074. 

c is approx 3e8 m/s. H0 is 70 km/s/Mpc = 70e3 m/s / (3.086e22 m/Mpc) ≈ 2.27e-18 s^-1. 

c/H0 ≈ 3e8 / 2.27e-18 ≈ 1.32e26 m. 

Then (2 / sqrt(0.7)) ≈ 2 / 0.8367 ≈ 2.39. 

So D_c ≈ 1.32e26 m * 2.39 * 1.074 ≈ 1.32e26 * 2.57 ≈ 3.39e26 m. 

Convert meters to Mpc: 3.39e26 m / 3.086e22 m/Mpc ≈ 1100 Mpc ≈ 1.1 Gpc. Wait, that's way less than the options given. But the options are 6 to 9 Gpc. So this approximation must be wrong. 

Wait, maybe the formula I used is incorrect. Let me check again. The formula I used might not be correct. Let's look it up. 

Upon checking, the correct formula for comoving distance in a flat universe (Omega_lambda + Omega_m =1) is:

D_c = (c / H0) * (1 / sqrt(Omega_m)) ) * ∫₀^z [1 / sqrt(1 + (Omega_lambda / Omega_m) * (1 + z')^3)] dz'

But I might have mixed up the terms. Alternatively, the integral is:

∫₀^z [1 / sqrt(Omega_m (1 + z')^3 + Omega_lambda)] dz'

Which can be expressed in terms of the hypergeometric function or special functions, but for an approximate calculation, perhaps using a numerical approximation would be better. 

Alternatively, use the relation between comoving distance and the age of the universe. The comoving distance is the integral of c/(a H) da from a=1 to a=1/(1+z). But again, this requires numerical integration. 

Alternatively, use an online calculator. But since I can't do that here, I'll try to estimate. 

For z=0.975, let's compute the integrand at several points and use Simpson's rule. 

The integrand is 1 / sqrt(0.3*(1+z')^3 + 0.7). Let's compute this at z'=0, 0.5, 0.75, 0.975. 

At z'=0: 1 / sqrt(0.3*1 + 0.7) = 1/sqrt(1) =1. 

At z'=0.5: 1 + z' =1.5. So 0.3*(1.5)^3 +0.7 =0.3*3.375 +0.7=1.0125 +0.7=1.7125. sqrt(1.7125)=1.3086. 1/1.3086≈0.764. 

At z'=0.75: 1 + z' =1.75. 0.3*(1.75)^3 +0.7=0.3*5.359 +0.7=1.6077 +0.7=2.3077. sqrt(2.3077)=1.519. 1/1.519≈0.658. 

At z'=0.975: 1 + z'=1.975. 0.3*(1.975)^3 +0.7=0.3*(7.656) +0.7≈2.2968 +0.7=3.0 (approx). sqrt(3.0)=1.732. 1/1.732≈0.577. 

Now, using Simpson's rule for four intervals (n=4, so 3 points between z'=0 and z'=0.975). The width of each interval is h=(0.975 -0)/4=0.24375. 

Simpson's rule formula: (h/3) [f0 + 4(f1 + f3) + 2f2 + fn] 

Here, f0=1, f1=0.764, f2=0.658, f3=0.577, fn=0.577 (since n=4, fn is the last point). 

So:

Integral ≈ (0.24375 /3) [1 + 4*(0.764 +0.577) +2*0.658 +0.577] 

Compute inside the brackets:

1 +4*(1.341) +1.316 +0.577 

1 +5.364 +1.316 +0.577 ≈ 8.257 

Multiply by (0.24375/3)=0.08125 

Integral ≈ 0.08125 *8.257≈0.671 

Then D_c = (c/H0) * integral ≈ (3e8 m/s / 2.27e-18 s^-1) *0.671 ≈1.32e26 m *0.671≈8.86e25 m. Convert to Mpc: 8.86e25 / 3.086e22 ≈286 Mpc. That's way too low. But the options are 6-9 Gpc. Clearly, this approach is flawed. 

Wait, maybe the issue is that the redshift z is not 0.975. Because the observed wavelength at 790 nm is much longer than the rest wavelength. Let's consider that the rest wavelength is, say, 400 nm (visible light), so the observed wavelength is 790 nm, which is a redshift of z=(790/400)-1=0.975. But then the comoving distance as calculated gives a much lower value than the options. So either I'm making a mistake in the calculation, or the approach is incorrect. 

Alternatively, perhaps the peak wavelength is not the rest wavelength. Maybe the peak is due to the thermal emission from the accretion disk of the quasar. The peak wavelength of thermal emission is given by the Wien's displacement law: lambda_peak * T = 2.898e-3 m·K. If the observed peak is 790 nm, then the temperature of the accretion disk would be T = 2.898e-3 / 790e-9 ≈ 3.68e3 K. That seems very low for an accretion disk, which is typically much hotter. The accretion disks around black holes are much hotter, like tens of thousands of Kelvin. So maybe that's not the case. 

Alternatively, perhaps the peak at 790 nm is not the rest wavelength but an emission line that's been redshifted. Suppose the quasar emits a line at a rest wavelength of 400 nm (blue), and due to redshift, it's observed at 790 nm. Then z=(790/400)-1=0.975. Then the comoving distance would be calculated for z=0.975. But as we saw earlier, that gives a comoving distance of about 286 Mpc, which is way less than the 6-9 Gpc options. 

Alternatively, perhaps the peak is at a longer wavelength due to some other effect. For example, if the quasar's emission is in the optical range, but due to relativistic effects, the observed wavelength is longer. But again, without knowing the rest wavelength, it's impossible to compute z. 

Wait, maybe the question is implying that the quasar is at a certain redshift where the peak is at 790 nm, and based on the cosmological parameters, we can compute the comoving distance. But unless we know the rest wavelength, we can't compute z. 

Alternatively, perhaps the question is a trick question, and the answer is based on the fact that the peak wavelength is part of the optical spectrum, implying that the quasar is at a moderate redshift, and the comoving distance in a flat Lambda-CDM universe at high z (say z~1) is around 1 Gpc. But the options are 6-9 Gpc. 

Wait, perhaps the quasar is at z=1. Let's compute the comoving distance for z=1. 

Using the same integral approach. 

At z=1, 1 + z=2. 

Integrand at z'=0: 1/sqrt(0.3 +0.7)=1. 

At z'=1: 1/sqrt(0.3*8 +0.7)=1/sqrt(2.4+0.7)=1/sqrt(3.1)≈0.566. 

Using Simpson's rule with two intervals (n=2). 

h=(1-0)/2=0.5. 

Integral ≈ (0.5/3)[f0 +4f1 +f2] 

f0=1, f1 at z'=0.5: 0.3*(1.5)^3 +0.7=0.3*3.375 +0.7=1.0125 +0.7=1.7125. sqrt(1.7125)=1.308. 1/1.308≈0.764. 

f2=0.566. 

So integral≈(0.5/3)[1 +4*0.764 +0.566] = (0.5/3)[1 +3.056 +0.566] = (0.5/3)(4.622)≈0.5*1.54≈0.77. 

Then D_c=(c/H0)*0.77. 

c/H0≈1.32e26 m. 

1.32e26 *0.77≈1.02e26 m. Convert to Mpc: 1.02e26 /3.086e22≈331 Mpc. Still way too low. 

Hmm. Maybe the redshift is higher. Let's try z=2. 

At z=2, 1+z=3. 

Integrand at z'=0:1. 

At z'=1:1/sqrt(0.3*8 +0.7)=1/sqrt(3.1)≈0.566. 

At z'=2:1/sqrt(0.3*27 +0.7)=1/sqrt(8.1 +0.7)=1/sqrt(8.8)=0.339. 

Using Simpson's rule with n=4 (h=0.5). 

Integral≈(0.5/3)[f0 +4(f1 +f3) +2f2 +f4]. 

f0=1, f1=0.764 (from z'=0.5), f2=0.566 (z'=1), f3=0.339 (z'=1.5), f4=0.339 (z'=2). 

Integral≈(0.5/3)[1 +4*(0.764 +0.339) +2*0.566 +0.339] 

= (0.5/3)[1 +4*(1.103) +1.132 +0.339] 

= (0.5/3)[1 +4.412 +1.132 +0.339] 

= (0.5/3)(6.883)≈0.5*2.294≈1.147 

D_c≈1.32e26 *1.147≈1.516e26 m≈1.516e26 /3.086e22≈492 Mpc. Still low. 

Wait, but the options are in Gpc. Maybe the redshift is much higher. Let's try z=5. 

At z=5, 1+z=6. 

Integrand at z'=0:1. 

At z'=1:0.566. 

At z'=2:0.339. 

At z'=3:1/sqrt(0.3*27 +0.7)=1/sqrt(8.1+0.7)=1/sqrt(8.8)=0.339. 

At z'=4:1/sqrt(0.3*64 +0.7)=1/sqrt(19.2 +0.7)=1/sqrt(19.9)=0.224. 

At z'=5:1/sqrt(0.3*125 +0.7)=1/sqrt(37.5 +0.7)=1/sqrt(38.2)=0.160. 

Using Simpson's rule with n=6 (h=0.8333). But this is getting complicated. Alternatively, maybe use an online approximation. 

Alternatively, use the approximation formula for the integral:

∫₀^z [1 / sqrt(a (1 + z')^3 + b)] dz' ≈ 2/(3*sqrt(a)) * [ (1 + z)^(3/2) -1 ] / sqrt(b + a*(1 + z)^3) 

But I'm not sure. Alternatively, use the fact that for high z, the integral approaches sqrt(1/(Omega_lambda)) * ln( (sqrt(Omega_lambda) + sqrt(Omega_m*(1+z)^3 + Omega_lambda)) / (sqrt(Omega_lambda) + sqrt(Omega_m + Omega_lambda)) ). 

But this is getting too involved. 

Alternatively, perhaps the answer is based on the fact that the Hubble length is about 1.32e26 m, which is 0.43 Gpc. Then, at z=2, the comoving distance is about 2 * Hubble length / (sqrt(1 - q0) ... ), but I'm not sure. 

Alternatively, maybe the answer is 8 Gpc (option D), as a rough estimate, but I need a better approach. 

Alternatively, perhaps the question is a trick and the answer is 0.7 Gpc (option A), but that seems too low. 

Wait, perhaps the key here is that the peak wavelength is 790 nm, which is in the near-infrared. The rest wavelength of the quasar's emission line is longer than that. For example, if the rest wavelength is 1.2 microns (1200 nm), then the observed wavelength is 790 nm, implying a redshift of z = (790/1200) -1 = 0.658 -1 = -0.342. That can't be. So perhaps the rest wavelength is 400 nm, giving z=0.975. But as we saw, that gives a comoving distance of about 286 Mpc, which is way too low. 

Wait, perhaps the question is referring to the fact that the flux drops at shorter wavelengths, implying that the peak is at the maximum of the Planck function for the blackbody spectrum of the quasar. But quasars don't have a blackbody spectrum; they are highly variable and non-thermal. 

Alternatively, maybe the peak wavelength is due to the thermal emission from a hotspot, but as before, the temperature would be too low. 

I'm stuck here. Maybe the answer is based on a standard cosmological calculation where for z=0.975, the comoving distance is around 8 Gpc. But how? 

Alternatively, perhaps the question is a trick and the peak wavelength is irrelevant, and the answer is based solely on the Hubble constant and the redshift inferred from the flux drop. But without the redshift, we can't compute the distance. 

Alternatively, perhaps the question is referring to the fact that the flux drops at shorter wavelengths, implying that the object is at a certain redshift where the rest wavelength is in the observed shorter wavelengths. But without knowing the rest wavelength, we can't compute z. 

Alternatively, maybe the question is implying that the flux drop indicates the peak is at 790 nm, which is the rest wavelength, and thus the quasar is at z=0. But that can't be, as comoving distance at z=0 is the current distance, which would be the Hubble distance. 

Wait, the Hubble distance is c/H0 ≈ 13.8 billion light-years ≈ 4.17 Gpc. The options are 6-9 Gpc. So if the quasar is at z=0.7, the comoving distance would be higher. Let's compute for z=0.7. 

Using Simpson's rule with n=4 (h=0.175). 

But this is getting too time-consuming. Maybe the correct approach is to recall that in a flat Lambda-CDM universe, the comoving distance for z=1 is about 2 Gpc. For higher z, it increases. Let's look for a rough estimate. 

Alternatively, use the approximation for comoving distance at high z: D_c ≈ (c/H0) * (1/(1 - q0)) * (z + sqrt(1 + 2 q0 z)) ), where q0 = -0.56. But I'm not sure. 

Alternatively, use the following approximate formula for comoving distance in a flat universe with matter and dark energy:

D_c ≈ (c / H0) * (2/(3 * sqrt(Ω_m))) * (1 + z) * [sqrt(1 + Ω_m (1 + z)^3 / Ω_lambda) -1] 

For Ω_m=0.3, Ω_lambda=0.7. Let's compute the term inside the brackets. 

sqrt(1 + 0.3*(1+z)^3 /0.7) = sqrt(1 + (3/7)*(1+z)^3). 

For z=0.975: 

(3/7)*(1.975)^3 ≈ (0.4286)*(7.656)≈3.28. 

So sqrt(1 +3.28)=sqrt(4.28)≈2.07. 

Then D_c ≈ (3e8 /2.27e-18) * (2/(3*sqrt(0.3))) *2.07 

Compute each part:

c/H0 ≈1.32e26 m. 

2/(3*sqrt(0.3))≈2/(3*0.5477)≈2/1.643≈1.217. 

Multiply all together: 1.32e26 *1.217 *2.07 ≈1.32e26 *2.53≈3.34e26 m. 

Convert to Gpc: 3.34e26 / 3.086e22 ≈1084 Mpc≈1.084 Gpc. Still too low. 

I think I'm missing something here. Maybe the question is not about the comoving distance but about the luminosity distance. Let's check. 

The luminosity distance D_L = (1+z) * D_c. For z=0.975, D_L≈1.975 *1.084≈2.13 Gpc. Still low. 

But the options are 6-9 Gpc. So perhaps the redshift is higher. Let's try z=2. 

For z=2: 

sqrt(1 +0.3*(3)^3 /0.7)=sqrt(1 +0.3*27 /0.7)=sqrt(1 +8.1/0.7)=sqrt(1 +11.57)=sqrt(12.57)=3.546. 

Then D_c≈1.32e26 * (2/(3*sqrt(0.3))) *3.546≈1.32e26 *1.217 *3.546≈1.32e26 *4.33≈5.72e26 m≈5.72e26 /3.086e22≈1850 Mpc≈1.85 Gpc. Still too low. 

Wait, maybe the cosmological parameters are different. Let's check the given H0=70 km/s/Mpc. The standard H0 for a flat universe with Ω_m=0.3 and Ω_lambda=0.7 is about H0≈70.4 km/s/Mpc, which matches the given value. So that's correct. 

Alternatively, perhaps the redshift is much higher, like z=5. Let's compute D_c for z=5. 

sqrt(1 +0.3*(6)^3 /0.7)=sqrt(1 +0.3*216 /0.7)=sqrt(1 +64.8/0.7)=sqrt(1 +92.57)=sqrt(93.57)=9.67. 

D_c≈1.32e26 *1.217 *9.67≈1.32e26 *11.76≈1.55e27 m≈1.55e27 /3.086e22≈5030 Mpc≈5.03 Gpc. Still low. 

Hmm. 

Alternatively, maybe the quasar's emission is in the near-infrared, and the observed peak is at 790 nm, which is 7.9 microns. The rest wavelength would be longer, say 79 microns, which is unrealistic. 

Alternatively, perhaps the rest wavelength is 400 nm (0.4 microns), observed at 790 nm (0.79 microns). Then z=(790/400)-1=0.975. Then the comoving distance is for z=0.975. But as before, that gives a comoving distance of ~1 Gpc. 

But the options are 6-9 Gpc. Maybe the question is using a different approach. 

Wait, perhaps the quasar's peak wavelength is in the near-infrared, and the flux drops at shorter wavelengths (visible), implying that the quasar's emission is redshifted into the infrared. But without knowing the rest wavelength, we can't compute z. 

Alternatively, perhaps the question is based on the fact that the peak wavelength corresponds to a certain rest wavelength, and the redshift can be estimated based on the observed flux drop. But I don't see how. 

Alternatively, perhaps the quasar's spectral energy distribution peaks at a certain wavelength, and the flux drops at shorter wavelengths. The peak wavelength is given, so perhaps the redshift is determined by the ratio of the observed peak to the rest peak. But without knowing the rest peak, it's impossible. 

Alternatively, maybe the question is a trick, and the answer is based on the fact that the Hubble constant is 70 km/s/Mpc, so the Hubble distance is c/H0 ≈4.24 Gpc. But the options are higher. 

Alternatively, perhaps the answer is based on the luminosity distance. For z=2, D_L≈2.13 Gpc. For z=3, D_L≈3.15 Gpc. Still low. 

Wait, maybe the redshift is 4. Let's compute for z=4. 

sqrt(1 +0.3*(5)^3 /0.7)=sqrt(1 +0.3*125 /0.7)=sqrt(1 +37.5 /0.7)=sqrt(1 +53.57)=sqrt(54.57)=7.387. 

D_c≈1.32e26 *1.217 *7.387≈1.32e26 *8.99≈1.18e27 m≈1.18e27 /3.086e22≈3820 Mpc≈3.82 Gpc. Still low. 

I'm really stuck here. Maybe the correct approach is to use the fact that the comoving distance for high z (z≈1) in a flat universe is about 2 Gpc, and for z≈2, it's about 4 Gpc. But the options are 6-9 Gpc, so maybe z≈4. Let's check. 

For z=4, D_c≈1.32e26 *1.217 *7.387≈1.32e26 *8.99≈1.18e27 m≈3.82 Gpc. Still low. 

Alternatively, perhaps the quasar is at a much higher redshift. Let's try z=5. 

sqrt(1 +0.3*6^3 /0.7)=sqrt(1 +0.3*216 /0.7)=sqrt(1 +64.8 /0.7)=sqrt(1 +92.57)=sqrt(93.57)=9.67. 

D_c≈1.32e26 *1.217 *9.67≈1.32e26 *11.76≈1.55e27 m≈1.55e27 /3.086e22≈5030 Mpc≈5.03 Gpc. Still low. 

Alternatively, maybe the cosmological parameters are different. Let's check for Ω_m=0.3 and Ω_lambda=0.7. 

Wait, perhaps the answer is based on the fact that the comoving distance for z=0.975 is approximately 8 Gpc. Let me check with a different approach. 

The integral ∫₀^z [1 / sqrt(0.3*(1+z')^3 +0.7)] dz' can be approximated by 2/(3*sqrt(Ω_m)) * (1 + z) for high z. But I'm not sure. 

Alternatively, use the approximation for large z: 

∫₀^z [1 / sqrt(Ω_m*(1+z')^3 + Ω_lambda)] dz' ≈ sqrt(1/Ω_lambda) * arcsinh( sqrt(Ω_lambda/Ω_m) (1+z) ) 

For z=0.975, Ω_lambda=0.7, Ω_m=0.3: 

sqrt(0.7/0.3)=sqrt(2.333)=1.5275. 

arcsinh(1.5275*(1+0.975))=arcsinh(1.5275*1.975)=arcsinh(3.012)≈1.493. 

Then D_c≈(c/H0) * 2/(3*sqrt(0.3)) *1.493≈1.32e26 *1.217 *1.493≈1.32e26 *1.817≈2.417e26 m≈2.417e26 /3.086e22≈7820 Mpc≈7.82 Gpc. 

That's close to option D (8 Gpc). 

So the answer is D. 8 Gpc.
</think>