If the deBroglie wavelength of an electron is equal to $$10^{-3}$$ times the wavelength of a photon of frequency $$6 \times 10^{14}$$ Hz, then the speed of electron is equal to: (Speed of light $$= 3 \times 10^8$$ m/s, Planck's constant $$= 6.63 \times 10^{-34}$$ J.s, Mass of electron $$= 9.1 \times 10^{-31}$$ kg)

We are told that the de Broglie wavelength of an electron is $$10^{-3}$$ times the wavelength of a photon whose frequency is $$6 \times 10^{14}\,\text{Hz}$$. First, we determine the wavelength of that photon.

The relation between the speed of light $$c$$, frequency $$\nu$$ and wavelength $$\lambda$$ of any photon is stated by the formula

$$c = \lambda_{\text{photon}} \,\nu_{\text{photon}}.$$

So, solving for the photon’s wavelength, we have

$$\lambda_{\text{photon}} = \frac{c}{\nu_{\text{photon}}}.$$

Substituting the given values $$c = 3 \times 10^{8}\,\text{m/s}$$ and $$\nu_{\text{photon}} = 6 \times 10^{14}\,\text{Hz}$$, we get

$$\lambda_{\text{photon}} = \frac{3 \times 10^{8}}{6 \times 10^{14}} = \frac{3}{6} \times 10^{8-14} = 0.5 \times 10^{-6}\,\text{m} = 5 \times 10^{-7}\,\text{m}.$$

Now, the electron’s de Broglie wavelength is stated to be $$10^{-3}$$ times this value, so

$$\lambda_{\text{electron}} = 10^{-3}\,\lambda_{\text{photon}} = 10^{-3}\,(5 \times 10^{-7}) = 5 \times 10^{-10}\,\text{m}.$$

For a material particle such as an electron, the de Broglie relation gives

$$\lambda_{\text{electron}} = \frac{h}{m_{\text{e}}\,v},$$

where $$h$$ is Planck’s constant, $$m_{\text{e}}$$ is the electron mass and $$v$$ is the electron’s speed. We rearrange this formula to solve for $$v$$:

$$v = \frac{h}{m_{\text{e}}\,\lambda_{\text{electron}}}.$$

Substituting the numerical values $$h = 6.63 \times 10^{-34}\,\text{J·s}, \qquad m_{\text{e}} = 9.1 \times 10^{-31}\,\text{kg}, \qquad \lambda_{\text{electron}} = 5 \times 10^{-10}\,\text{m},$$ we obtain

$$v = \frac{6.63 \times 10^{-34}} {(9.1 \times 10^{-31})(5 \times 10^{-10})}.$$

We multiply the two numbers in the denominator first:

$$(9.1 \times 10^{-31})(5 \times 10^{-10}) = 9.1 \times 5 \times 10^{-31-10} = 45.5 \times 10^{-41} = 4.55 \times 10^{-40}.$$

Now we divide:

$$v = \frac{6.63 \times 10^{-34}}{4.55 \times 10^{-40}} = \left(\frac{6.63}{4.55}\right)\times 10^{-34 - (-40)} = 1.457 \times 10^{6}\,\text{m/s}.$$

Rounding to two significant figures, the speed is

$$v \approx 1.45 \times 10^{6}\,\text{m/s}.$$

Hence, the correct answer is Option D.