If an electron has a wavelength of $0.123 \mathrm {~nm}$, what is its kinetic energy, in electron-volts? This energy is not in the relativistic region. ( $m$ electron $= 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , h = 6.626$ $\times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s }$ )
A) $99.5 \mathrm { eV }$
B) $60.2 \mathrm { eV }$
C) $80.4 \mathrm { eV }$
D) $19.8 \mathrm { eV }$
E) $124 \mathrm { eV }$

Question

Quizplus · Accepted Answer

The de Broglie wavelength formula $\lambda = \frac{h}{mv}$ can be rearranged to find the velocity $v$. Using $v$, the kinetic energy $KE = \frac{1}{2}mv^2$ can be calculated. Converting this energy to electron-volts gives approximately 99.5 eV.

If an Electron Has a Wavelength Of 0.123 nm0.123 \mathrm {~nm}0.123 nm , What Is Its Kinetic Energy, in Electron-Volts? This Energy

If an Electron Has a Wavelength Of $0.123 \mathrm {~nm}$ , What Is Its Kinetic Energy, in Electron-Volts? This Energy