Final answer:
The de Broglie wavelength of an electron accelerated through a potential of 100 volts can be found by using the de Broglie equation \(\lambda = h/p\), with the kinetic energy from the potential difference and solving for the momentum of the electron.
Step-by-step explanation:
The question asks us to calculate the de Broglie wavelength of an electron accelerated through a potential of 100 volts. This involves using the formula for de Broglie wavelength, which is \(\lambda = h/p\), where \(h\) is Planck's constant and \(p\) is the momentum of the electron. The kinetic energy (KE) gained by the electron when it is accelerated through a potential difference (V) is given by \(KE = eV\), where \(e\) is the elementary charge. The momentum \(p\) of the electron can be calculated as \(p = \sqrt{2m_eKE}\), where \(m_e\) is the rest mass of the electron.
By substituting the KE with \(eV\) and then using the momentum formula, we can derive the de Broglie wavelength for the electron as follows:
\(\lambda = \frac{h}{\sqrt{2m_eeV}}\)
Now we plug in the values \(h = 4.135667696 \times 10^{-15} \text{eV}\cdot\text{s}\), \(m_e = 9.10938356 \times 10^{-31} \text{kg}\), \(e = 1.602176634 \times 10^{-19} \text{C}\) and \(V = 100 \text{V}\) into the equation to find the de Broglie wavelength in angstroms (Å).