Answer
See the answer below.
Work Step by Step
a) $c=\lambda\times\nu$
$\lambda=2.998\times10^8\ m/s\div 850 \times10^{6}\ 1/s=0.35\ m$
b)
Per photon:
$E=h\times\nu$
$E=6.626\times10^{-34}\ J.s\times 850\times10^{6}\ 1/s=5.63\times10^{-25}\ J$
Per mole:
$E=5.63\times10^{-25}\ J\times6.022\times10^{23}\ 1/mol=0.34\ J/mol$
c) $c=\lambda\times\nu$
$\nu=2.998\times10^8\ m/s\div 420 \times10^{-9}\ m=7.14\times10^{14}\ 1/s$
Per photon:
$E=h\times\nu$
$E=6.626\times10^{-34}\ J.s\times 7.14\times10^{14}\ 1/s=4.73\times10^{-19}\ J$
Per mole:
$E=4.73\times10^{-19}\ J\times6.022\times10^{23}\ 1/mol=284.8\ kJ/mol$
Which is much bigger than the cell phone signals.
d) $284.8\times10^3/0.34=840,000$ times greater energy per mole of violet light.
