 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.^[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 $dt+M{r}^2$, not $(dt{)}^2+M^2{r}^2$. 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 $s_{\nu }=F_{\nu }/N_{\mathrm{pix}}$. The ideal source and background electron counts are $S=K{s}_{\nu }$ and $B=K{b}_{\nu }{\Omega }_{\mathrm{pix}}$. 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. ThereforeThus 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 $b_{\nu }{\Omega }_{\mathrm{pix}}=1.75\times 10^{-32}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{H}{\mathrm{z}}^{-1}$. Multiplying arcsecond-squared pixel area by brightness per arcsecond squared already gives a flux: no additional steradian conversion belongs in this product. Rearranging givesUsing $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ yields

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. 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.^[ref]

A measured silicon pair-creation energy $w\simeq 3.65\mathrm{eV}$ gives $N_e=E/w\simeq 2.74\times 10^3$ 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. ThenThe Fano-limited X-ray CCD energy resolution using the representative silicon value $F\simeq 0.12$ gives an ideal Fano-limited $\Delta E_{\mathrm{FWHM}}\simeq 0.16\mathrm{keV}$ at $10\mathrm{keV}$, or resolving power $E/\Delta E\simeq 64$. 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.^[ref]