If the de Broglie wavelength of an electron is 2.4 ?m, what is the speed of the electron? $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.$ $\left. h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$ A)3.0 × 10² m/s B)2.5 × 10⁵ m/s C)1.7 × 10³ m/s D)8.3 × 10⁶ m/s

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Quizplus · Accepted Answer

The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. The momentum of an electron is given by $p = mv$, where $m$ is the mass of the electron and $v$ is its speed. Substituting these equations into the first equation, we get $\lambda = \frac{h}{mv}$. Solving for $v$, we get $v = \frac{h}{m\lambda}$. Plugging in the given values, we get $v = \frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{(9.11 \times 10^{-31} \mathrm{~kg})(2.4 \times 10^{-6} \mathrm{~m})} = 3.0 \times 10^2 \mathrm{~m/s}$.

If the De Broglie Wavelength of an Electron Is 2 (melectron =9.11×10−31 kg,\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.(melectron ​=9.11×10−31 kg,

If the De Broglie Wavelength of an Electron Is 2 $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.$