First, determine the wavelength of the photon. To do so, apply the formula of energy of photon.
`E = hf`
Since the frequency of light is `f=c/lambda`, the formula of photon's energy can be re-written as:
Then, isolate the wavelength of the photon.
Plugging in the values, the formula becomes:
`lambda = ((6.63 xx10^(-34)J*s )(3xx10^8 m/s))/(3.4eV*(1.60xx10^(-19)J)/(1eV))`
So, the wavelength of the photon is `3.65625 xx10^(-7)` m.
Next, consider the formula for DeBroglie wavelength.
Then, isolate the speed of the particle.
Since it is given that the wavelength of the photon is the same as the DeBroglie wavelength of the electron, plug in `lambda=3.65625xx10^(-7).' m. Also, plug in the value of the Planck's constant and the mass of electron.
Therefore, the speed of the electron is 1990.47 m/s.