In astronomical observations, signals observed from the distant stars are generally weak.

Solution:

From the expression of energy of one photon $(E)$,

$E=\frac{\text { he }}{\lambda}$

Where, $\lambda=$ wavelength of

radiation $h=$ Planck's

constant $c=$ velocity of

radiation

Substituting the values in the given expression of $E$ :

$=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}=3.313 \times 10^{-19} \mathrm{~J}$

Energy of one photon $=3.313 \times 10^{-19} \mathrm{~J}$

Number of photons received with $3.15 \times 10^{-18} \mathrm{~J}$ energy

$=\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}$

$=9.5$

$\approx 10$