After an electron has been accelerated through a potential difference of $0.15 \mathrm { kV }$, what is its de Broglie wavelength? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$
A) $1.0 \mathrm {~nm}$
B) $1.0 \mathrm {~mm}$
C) $0.10 \mathrm {~nm}$
D) $1.0 \mathrm {~cm}$

Question

Quizplus · Accepted Answer

The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum. For an electron accelerated through a potential difference $V$, $p = \sqrt{2m_e e V}$. Substituting the given values, $\lambda \approx 0.10 \mathrm{~nm}$.

After an Electron Has Been Accelerated Through a Potential Difference 0.15kV0.15 \mathrm { kV }0.15kV

After an Electron Has Been Accelerated Through a Potential Difference $0.15 \mathrm { kV }$