Step 1 : Understanding the Question:
This problem asks us to determine the wave-like wavelength, known as the de-Broglie wavelength, associated with an electron that has been accelerated from a state of rest through an electric potential difference of 100 volts. According to quantum mechanics, moving particles exhibit wave-like characteristics, and their wavelengths are inversely proportional to their momentum. We need to relate the electric potential to the kinetic energy, then to the momentum, and finally to the de-Broglie wavelength of the electron.
Step 2 : Key Formulas and Approach:
The de-Broglie relationship states that the wavelength $\lambda$ of a moving particle is given by:
\[ \lambda = \frac{h}{p} \]
Where $h$ is Planck's constant and $p$ is the momentum of the particle.
The kinetic energy $K$ of a charged particle accelerated through a potential difference $V$ is given by:
\[ K = q V \]
The relationship between kinetic energy and momentum is:
\[ p = \sqrt{2 m K} = \sqrt{2 m q V} \]
Substituting this into the de-Broglie equation gives:
\[ \lambda = \frac{h}{\sqrt{2 m q V}} \]
For an electron, we can substitute the values of the constant physical quantities to arrive at a simplified practical formula.
Step 3 : Detailed Explanation:
We start by substituting the physical constants for an electron into the general wavelength formula. The mass of an electron is $m \approx 9.1 \times 10^{-31}\text{ kg}$, the charge is $q = e \approx 1.6 \times 10^{-19}\text{ C}$, and Planck's constant is $h \approx 6.626 \times 10^{-34}\text{ J s}$.
Substituting these constant values into $\lambda = \frac{h}{\sqrt{2 m e V}}$ yields a highly simplified formula specifically for electrons: $\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}$, where $1\text{ \AA} = 10^{-10}\text{ m}$.
Next, we identify the given accelerating potential difference from the problem, which is $V = 100\text{ V}$.
We substitute $V = 100$ into our simplified formula: $\lambda = \frac{12.27}{\sqrt{100}} \text{ \AA}$.
Since the square root of 100 is exactly 10, the expression simplifies to: $\lambda = \frac{12.27}{10} \text{ \AA}$.
Performing this division shifts the decimal point one place to the left, resulting in $\lambda = 1.227 \text{ \AA}$.
Step 4 : Final Answer:
The de-Broglie wavelength of the accelerated electron is approximately $1.227 \text{ \AA}$, which corresponds to Option (A).