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Step-by-Step Solution
Step 1: Determine the wavelength of the photon
The frequency of the photon is given as $6\times 10^{14}\,\text{Hz}$. The wavelength $ (\lambda_{\text{photon}}) $ of a photon is related to its frequency $ (\nu) $ and speed of light $ (c) $ by the relation:
$ \lambda_{\text{photon}} = \frac{c}{\nu} $
Here, $ c = 3 \times 10^8 \,\text{m/s}$ and $ \nu = 6 \times 10^{14}\,\text{Hz}$. Substituting the values, we get:
$ \lambda_{\text{photon}} = \frac{3 \times 10^8}{6 \times 10^{14}}
= 5 \times 10^{-7}\,\text{m}
\,.$
Step 2: Express the de Broglie wavelength of the electron
The de Broglie wavelength $ (\lambda_e) $ of an electron of mass $ m $ moving with speed $ v $ is given by:
$ \lambda_e = \frac{h}{m\,v} \,,$
where $ h $ is Planck's constant ($6.63 \times 10^{-34}\,\text{J}\cdot\text{s}$). We are given that:
$ \lambda_e = 10^{-3} \, \lambda_{\text{photon}} \,.
$
Step 3: Set up the equality for the wavelengths
According to the problem statement:
$ \frac{h}{m\,v} = 10^{-3} \left( \frac{c}{\nu} \right) \,.
$
Substituting $ \lambda_{\text{photon}} = \frac{3 \times 10^8}{6 \times 10^{14}} $ into the above relation, we get:
$ \frac{h}{m\,v} = 10^{-3}
\left(\frac{3 \times 10^8}{6 \times 10^{14}}\right) \,.
$
Step 4: Solve for the speed of the electron
Rearrange to find $ v $:
$ v = \frac{h \times \nu \times 10^{3}}{m \times c} \,
$
However, the simplest way is to use the given form directly in numeric substitution:
$ v = \frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^5} \,.
Notice that $10^5$ in the denominator comes from multiplying $3 \times 10^8$ by $10^{-3}$, giving $3 \times 10^5$ effectively. Performing the calculation:
$ v = 1.45 \times 10^6 \,\text{m/s} \,.
Answer
The speed of the electron is $1.45 \times 10^6\,\text{m/s}$.
