What’s the energy of a photon whose wavelength is $2.07\ \mathrm{nm}$?
A) $60\ \mathrm{eV}$
B) $600\ \mathrm{eV}$
C) $960\ \mathrm{eV}$
D) $6000\ \mathrm{eV}$
E) $9600\ \mathrm{eV}$
A) $60\ \mathrm{eV}$
B) $600\ \mathrm{eV}$
C) $960\ \mathrm{eV}$
D) $6000\ \mathrm{eV}$
E) $9600\ \mathrm{eV}$
Combining the equation
$$E=hf$$
with
$$f=\displaystyle \frac{c}{\lambda }$$
gives us
$$E=\frac{hc}{\lambda }$$
$$E=\frac{\left(4.14\times 10^{-15}\ \mathrm{eV\cdot s}\right)\left(3.00\times 10^{8}\ \mathrm{\displaystyle \frac{m}{s}}\right)}{2.07\times 10^{-9}\ \mathrm{m}}$$
$$E=600\ \mathrm{eV}$$