Solution
= Solution
Each <photon> has <energy> $hf$, giving the <photon emission rate from monochromatic power> and <wavelength>
$$
\boxed{\dot N=\frac{P}{hf}=8.38\times10^{22}\,\mathrm{s^{-1}},\qquad
\lambda=\frac cf=0.167\,\mathrm m.}
$$
Here $P$ is radiated signal <power>, not the phone's total electrical consumption.