Question

Determine the de Broglie wavelength of an electron accelerated through \(100\,V\).

• A. \(0.1227\,\text{\AA}\)
• B. \(1.227\,\text{\AA}\)
• C. \(12.27\,\text{\AA}\)
• D. \(0.01227\,\text{\AA}\)

Hint:

Start from the energy gained by an electron, $K=eV$, and use it in $p=\sqrt{2m_eK}$. Once the momentum is known, apply $\lambda=h/p$ and convert metres to angstroms.

Answer 1

The Correct Option is B

Concept: According to the de Broglie hypothesis, a moving particle exhibits wave-like properties. The wavelength associated with a charged particle accelerated through a potential difference \(V\) is given by \[ \lambda = \frac{h}{p} \] For electrons accelerated through a potential \(V\), this simplifies to \[ \lambda = \frac{12.27}{\sqrt{V}} \; \text{\AA} \] where \(V\) is in volts.

Step 1: Write the de Broglie wavelength formula for an electron. \[ \lambda = \frac{12.27}{\sqrt{V}} \; \text{\AA} \]

Step 2: Substitute the given potential difference. \[ V = 100 \] \[ \lambda = \frac{12.27}{\sqrt{100}} \]

Step 3: Evaluate the expression. \[ \lambda = \frac{12.27}{10} \] \[ \lambda = 1.227 \; \text{\AA} \]

Answer 2

Concept:

• Derive the wavelength from the energy gained by the electron instead of using the ready-made electron-voltage shortcut.
• Use $eV=\dfrac{p^2}{2m_e}$ together with $\lambda=\dfrac{h}{p}$.

Step 1: Convert the gained energy into joules.
The electron gains $100\,\text{eV}$ of kinetic energy.
$K=100(1.602\times10^{-19})=1.602\times10^{-17}\,\text{J}$

Step 2: Calculate the electron momentum.
$p=\sqrt{2m_eK}$
$=\sqrt{2(9.11\times10^{-31})(1.602\times10^{-17})}$
$=5.40\times10^{-24}\,\text{kg m s}^{-1}$

Step 3: Apply the de Broglie relation.
$\lambda=\dfrac{h}{p}=\dfrac{6.626\times10^{-34}}{5.40\times10^{-24}}$
$\lambda=1.227\times10^{-10}\,\text{m}$

Step 4: Convert metres to angstroms.
Since $1\,\text{angstrom}=10^{-10}\,\text{m}$,
$\lambda=1.227\,\text{angstrom}$

Final Answer: $1.227\,\text{angstrom}$, option B.