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Step-by-Step Solution
Step 1: Write down the known information
We are given that an electron is accelerated from rest through a potential difference $V$ volts. Its de-Broglie wavelength is $\lambda = 1.227 \times 10^{-2}\,\text{nm}$.
Step 2: Convert the wavelength to meters (if needed for clarity)
1 nm = $10^{-9}$ m
$\lambda = 1.227 \times 10^{-2} \,\text{nm} = 1.227 \times 10^{-2} \times 10^{-9}\,\text{m} = 1.227 \times 10^{-11}\,\text{m}.$
Step 3: Use the approximate formula for the de-Broglie wavelength of an electron
For an electron accelerated by a potential $V$, a commonly used approximate relation (in Å) is:
$ \lambda\,(\text{Å}) = \frac{12.27}{\sqrt{V(\text{volts})}}. $
Note that 1 Å = $10^{-10}$ m.
Step 4: Convert the given wavelength to Angstroms (Å)
$\lambda = 1.227 \times 10^{-11}\,\text{m} = 0.1227 \times 10^{-10}\,\text{m} = 0.1227\,\text{Å}.$
However, from the given relation, one may directly use the ratio form. Let us equate the formula carefully.
Step 5: Plug in the numerical values and solve for $\sqrt{V}$
According to the formula,
$ \lambda (\text{Å}) = \frac{12.27}{\sqrt{V}}.
$
Given $\lambda (\text{Å}) = 0.1227$, we have
$ 0.1227 = \frac{12.27}{\sqrt{V}}.
$
Rearranging,
$ \sqrt{V} = \frac{12.27}{0.1227} = 100.
$
Step 6: Calculate the potential difference $V$
Since $\sqrt{V} = 100,$ we get
$ V = (\sqrt{V})^2 = 100^2 = 10^4 \,\text{volts}.
Step 7: Conclude the answer
The potential difference through which the electron is accelerated (to have the given de-Broglie wavelength) is $10^4$ volts.
