The frequency of the [tex]\lambda_2 = 622 nm = 622 \cdot 10^{-9} m[/tex] wavelength photon is given by
[tex]f_2 = \frac{c}{\lambda_2}= \frac{3 \cdot 10^8 m/s}{622 \cdot 10^{-9} m}=4.82 \cdot 10^{14} Hz [/tex]
where c is the speed of light.
The energy of this photon is
[tex]E_2=hf_2 = (6.6 \cdot 10^{-34}Js)(4.82 \cdot 10^{14}Hz)=3.18 \cdot 10^{-19} J[/tex]
where h is the Planck constant.
The energy of the first photon is twice that of the second photon, so
[tex]E_1 = 2 E_2 = 2 \cdot 3.18 \cdot 10^{-19}J =6.36 \cdot 10^{-19} J[/tex]
And so now by using again the relationship betwen energy and frequency, we can find the frequency of the first photon:
[tex]f_1 = \frac{E_1}{h}= \frac{6.36 \cdot 10^{-19} J}{6.6 \cdot 10^{-34}Js}=9.64 \cdot 10^{14}Hz [/tex]
and its wavelength is
[tex]\lambda_1 = \frac{c}{f_1}= \frac{3 \cdot 10^8 m/s}{9.64 \cdot 10^{14}Hz} =3.11 \cdot 10^{-7}m = 311 nm [/tex]
So, we see that the wavelength of the first photon is exactly half of the wavelength of the second photon (622 nm).
