In a computer monitor,electrons approach the screen at 1.20 × 10⁸ m/s.What is the de Broglie wavelength of these electrons? Note: the mass of electrons is 9.109 × 10^-³¹ kg;and use the relativistic momentum in your calculation. A)4.31 × 10^-¹² m B)5.56 × 10^-¹² m C)6.07 × 10^-¹² m D)6.62 × 10^-¹² m E)7.85 × 10^-¹² m

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Quizplus · Accepted Answer

The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$, where $h$ is Planck's constant ($6.626 \times 10^{-34}$ m$^2$kg/s) and $p$ is the relativistic momentum. The relativistic momentum can be calculated using the formula $p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $m$ is the mass of the electron ($9.109 \times 10^{-31}$ kg), $v$ is the velocity of the electron ($1.20 \times 10^8$ m/s), and $c$ is the speed of light ($3.00 \times 10^8$ m/s). After calculating $p$, substitute it into the de Broglie wavelength formula to find $\lambda$. The correct calculation leads to the wavelength closest to the value given in option B.

In a Computer Monitor,electrons Approach the Screen at 1