Question

part a
the de broglie wavelength of an electron with a velocity of 6.10 × 10⁶ m/s is ______ m. the mass of the electron is 9.11 × 10⁻²⁸ g.
○ 8.39 × 10¹²
○ 8.39 × 10⁹
○ 1.19 × 10⁻¹⁶
○ 1.19 × 10⁻¹³
○ 1.19 × 10⁻¹⁰
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Explanation:

Step1: Recall de Broglie wavelength formula

The de Broglie wavelength formula is \(\lambda=\frac{h}{mv}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant), \(m\) is the mass of the particle, and \(v\) is the velocity of the particle.

First, convert the mass of the electron from grams to kilograms. Given \(m = 9.11\times10^{-28}\space g\). Since \(1\space kg= 1000\space g\), so \(m=9.11\times10^{-28}\space g\times\frac{1\space kg}{1000\space g}=9.11\times10^{-31}\space kg\)

Step2: Substitute values into the formula

We know \(v = 6.10\times10^{6}\space m/s\), \(h = 6.626\times10^{-34}\space J\cdot s\) and \(m = 9.11\times10^{-31}\space kg\)

Substitute these values into \(\lambda=\frac{h}{mv}\):

\[

$$\begin{align*} \lambda&=\frac{6.626\times 10^{-34}\space J\cdot s}{(9.11\times 10^{-31}\space kg)\times(6.10\times 10^{6}\space m/s)}\\ &=\frac{6.626\times 10^{-34}}{9.11\times6.10\times 10^{-31 + 6}}\\ &=\frac{6.626\times 10^{-34}}{55.571\times 10^{-25}}\\ &=\frac{6.626}{55.571}\times10^{-34 + 25}\\ &\approx0.119\times 10^{-9}\\ &= 1.19\times 10^{-10}\space m \end{align*}$$

\]

Answer:

\(1.19\times 10^{-10}\) (corresponding to the option \(1.19\times 10^{-10}\))