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:

`E=(hc)/lambda`

Then, isolate the wavelength of the photon.

`lambda= (hc)/E`

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))`

`lambda=3.65625 xx10^(-7)`

So, the wavelength of the photon is `3.65625 xx10^(-7)` m.

Next, consider the formula for DeBroglie wavelength.

`lambda=h/(mv)`

Then, isolate the speed of the particle.

`v=h/(mlambda)`

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.

`v=(6.63xx10^(-34) J*s)/((9.11xx10^(-31)kg)(3.65625xx10^(-7)m))`

`v=1990.47 m/s`

**Therefore, the speed of the electron is 1990.47 m/s.**

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