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# Find the energy of a photon in electron volt of microwaves having wavelength $3mm.(1\: eV = 1.6 \times 10^{-19}J)$

$(a) 41.4 \times 10^{-2}\; eV\quad (b) 4.14 \times 10^{-4}\; eV$
$(c) 4.14 \times 10^{-2}\; eV\quad (b) 41.4 \times 10^{-4}\; eV$

Energy = $\large\frac{hc}{\lambda}$ $=\large\frac{ 6.6 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-3}}$$= \large\frac{6.6 \times 10^{-23} J}{ 1.6 \times 10^{-19}\: eV}$$ = 4.14 \times 10^{-4}\: eV$
Ans : (b)