Solution
= Solution
In a <CCD>, absorbed optical <photons> create photoelectrons collected in electrostatic wells beneath gate electrodes. During readout, clock voltages move the charge packets row by row into a serial register; serial clocks move them one <pixel> at a time to an output node. The node converts charge to a voltage, approximately $\Delta V=Q/C$, which is amplified and digitized. <pixel> addresses are recovered from the row and serial-clock sequence. Resetting and sampling the output between packets establishes the baseline. Efficient transfer preserves charge; trapping or incomplete transfer produces loss and trailing. This is <CCD clocked charge transfer>, not independent wiring of an <amplifier> to every imaging <pixel>.<ASCA Technical Description: SIS by NASA HEASARC>[§7, SIS: charge generation, electrode-clock transfer and preamplifier readout]{ref}
<dark current> is thermally generated charge accumulated even without illumination. If its mean rate is $d$ <electrons> per <pixel> per second, a single exposure adds mean $dt$ and Poisson <variance> $dt$. Cooling the detector and using low-defect material suppresses this generation. A dark <calibration> removes the reproducible mean pattern, but cannot remove the dark <electrons>' random <shot noise> in the science exposure. <read noise> comes from measuring the output packet and resetting the sense node, <amplifiers> and electronics; write its rms as $r$ <electrons> per read. <correlated double sampling> reduces reset and correlated offsets by subtracting the baseline measurement from the signal measurement. Low-noise electronics and an appropriate slower readout reduce the remaining contribution. The <CCD noise budget> for $M$ separate exposures combined over total time $t$ contains independent dark and read contributions $dt+Mr^2$, not $(dt)^2+M^2r^2$. Long exposures reduce repeated <read noise>, but saturation, cosmic rays and tracking requirements constrain the exposure length. On-chip binning can reduce read-noise penalties for a fixed binned area at the cost of angular sampling.
For the source and sky calculation, define the count conversion factor $K=At\Delta\nu/(h\nu)$ and the source flux per <pixel> $s_\nu=F_\nu/N_{\rm pix}$. The ideal source and background <electron> counts are $S=Ks_\nu$ and $B=Kb_\nu\Omega_{\rm pix}$. These assume unit throughput and <quantum efficiency>, approximately constant quantities across the band and a well-determined mean sky level from many background <pixels>. Source and sky <photons> have independent Poisson fluctuations, so the <variance> after subtracting that mean sky is $S+B$. Subtracting a background removes its mean, not its <photon> noise. Therefore
$$
R=\frac S{\sqrt{S+B}}=\frac{F_\nu}{N_{\rm pix}}\left[\frac{At\Delta\nu}{h\nu(F_\nu/N_{\rm pix}+b_\nu\Omega_{\rm pix})}\right]^{1/2}.
$$
Thus \b[the required expression follows directly from adding photon-count <variances>]. Finite sky-estimation noise, <dark current> and <read noise> would add further <variance>. A nonunit common efficiency replaces $A$ by the effective collecting area; a nonflat spectrum requires the corresponding <frequency> integral. For example, estimating the sky from one equally exposed background <pixel> would add an extra $B$, whereas a large sky region makes that extra term small.
With $A=\pi(1\,\mathrm m)^2$ and the supplied <photon> <frequency>, the sky flux per <pixel> is $b_\nu\Omega_{\rm pix}=1.75\times10^{-32}\,\mathrm{W\,m^{-2}\,Hz^{-1}}$. Multiplying arcsecond-squared <pixel> area by brightness per arcsecond squared already gives a flux: no additional steradian conversion belongs in this product. Rearranging gives
$$
\frac t{R^2}=\frac{h\nu}{A\Delta\nu}\frac{F_\nu/N_{\rm pix}+b_\nu\Omega_{\rm pix}}{(F_\nu/N_{\rm pix})^2}.
$$
Using $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$ yields
\Table[
|| $F_\nu\ (\mathrm{W\,m^{-2}\,Hz^{-1}})$
|| Source rate per <pixel> $\mathrm{s^{-1}}$
|| Sky rate per <pixel> $\mathrm{s^{-1}}$
|| $t/R^2\ (\mathrm s)$
| $10^{-29}$
| $98.8$
| $13.8$
| \b[$0.0115$]
| $10^{-30}$
| $9.88$
| $13.8$
| \b[$0.243$]
| $10^{-31}$
| $0.988$
| $13.8$
| \b[$15.2$]
]
{title=Ideal per-pixel source and sky rates and exposure divided by squared signal-to-noise ratio}
The first case is mainly source-shot-noise limited, while the last is strongly sky limited. At high source brightness $t/R^2\propto F_\nu^{-1}$; at low source brightness it approaches $F_\nu^{-2}$. Reducing source flux by ten can therefore require nearly a hundred times the exposure once sky noise dominates. These are per-pixel values for the ideal model, not guaranteed integration times including <calibration>, cosmic rays or <read noise>.
Increasing the diameter increases both source and sky <photon> rates as $D^2$ for fixed angular <pixels>. In this ideal limit the required exposure decreases as $D^{-2}$, whereas the stated <telescope> cost increases as $D^{2.5}$: collecting area grows only as cost to the power $0.8$. A larger ground <telescope> does not remove atmospheric <photons> or recover arbitrary wavefront information lost in a long seeing-limited exposure. Space observations avoid atmospheric airglow, weather and <atmospheric turbulence>, and offer stable high-resolution imaging and access to very faint structure. They still have zodiacal light, instrumental backgrounds and their own <calibration> limits. <adaptive optics> can provide excellent ground-based resolution in suitable fields, but its guide-star, <wavelength> and field-dependent limitations do not remove the need for space-based continuum imaging. \b[The need is for complementary low-background, stable and high-resolution measurements; the cost exponent alone is not a proof that every imaging programme is cheaper in space.]
For X rays, one fully absorbed <photon> produces many <electron-hole pairs> in <silicon>, rather than one optical photoelectron count. At moderate <surface brightness>, the <probability> of two <photons> landing in the same event island during one <CCD> frame is small. The charge packet or sum over neighbouring split-event <pixels> then supplies a pulse height for each <photon>, while its <pixel> position supplies the image coordinate. Calibrating pulse height against energy produces an event list from which a spectrum can be accumulated for any spatial region. <photon pile-up> at larger occupancies merges <photon> energies and corrupts both spectra and event grades, so fast enough frames and event-pattern screening are essential.<ASCA Technical Description: SIS by NASA HEASARC>[§7, SIS: silicon electron-hole yield, charge collection and imaging spectroscopy]{ref}
A measured <silicon> pair-creation energy $w\simeq3.65\,\mathrm{eV}$ gives $N_e=E/w\simeq2.74\times10^3$ <electrons> for a $10\,\mathrm{keV}$ <photon>. Let $F$ be the <Fano factor> of pair production, $n_r$ the number of independently read <pixels> in the event sum, and $r$ the rms <read noise> per <pixel>. Then
$$
\sigma_E=w\sqrt{F\frac Ew+n_rr^2+\sigma_d^2+\cdots},\qquad \Delta E_{\rm FWHM}=2\sqrt{2\ln2}\,\sigma_E.
$$
The <Fano-limited X-ray CCD energy resolution> using the representative <silicon> value $F\simeq0.12$ gives an ideal Fano-limited \b[$\Delta E_{\rm FWHM}\simeq0.16\,\mathrm{keV}$ at $10\,\mathrm{keV}$], or resolving power $E/\Delta E\simeq64$. Finite <read noise>, charge loss and event summing broaden this; an order-of-magnitude practical estimate is a few tenths of a keV, but the particular detector's value is not determined by the given optical-pixel parameters. These $w$ and $F$ values are external detector properties, not supplied question data. Treating every pair as independent Poisson production instead would give $0.19\,\mathrm{keV}$ rms, or $0.45\,\mathrm{keV}$ FWHM; stating which width is meant avoids conflating those estimates. Radiation-induced charge-transfer inefficiency requires additional gain and spectral-response <calibration>.<The ASCA Data Reduction Guide by NASA HEASARC>[§4.7, non-uniform CTI: silicon yield, Fano factor near 0.12 and FWHM resolution formula]{ref}