Question

What is the de-Broglie wavelength of an electron accelerated from rest through a potential difference of 100 volts?

• A. None of these
• B. 122A
• C. 1.2A
• D. 0.122A

Answer 1

Answer: (C)

1.2A

Answer 2

The de-Broglie wavelength ($\lambda$) of an electron accelerated from rest through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2m_e eV}}$ where:

• $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J s}$)
• $m_e$ is the mass of an electron ($9.109 \times 10^{-31} \text{ kg}$)
• $e$ is the charge of an electron ($1.602 \times 10^{-19} \text{ C}$)
• $V$ is the potential difference in volts

Substituting the values of the constants, this formula can be simplified for an electron to: $\lambda (\text{in Angstroms}) = \frac{12.27}{\sqrt{V}}$ Given the potential difference $V = 100 \text{ volts}$. Substitute the value of $V$ into the simplified formula: $\lambda = \frac{12.27}{\sqrt{100}}$ $\lambda = \frac{12.27}{10}$ $\lambda = 1.227 \text{ Å}$ The calculated de-Broglie wavelength $1.227 \text{ Å}$ is closest to $1.2 \text{ Å}$.