Question

The de-Broglie wavelength of an electron moving with a velocity of \(10^7\) m/s is:

• A. \(7.3 \times 10^{-11}\) m
• B. \(1.3 \times 10^{-11}\) m
• C. \(7.3 \times 10^{-7}\) m
• D. \(3.1 \times 10^{-7}\) m

Hint:

For calculations involving fundamental constants, it's often useful to know approximate ratios. For instance, \(h/m_e \approx 7.27 \times 10^{-4}\). This can sometimes speed up calculations. Also, always double-check the powers of ten.

Answer 1

The Correct Option is A

Step 1: State the de-Broglie wavelength equation. The de-Broglie wavelength (\(\lambda\)) is calculated as \(\lambda = \frac{h}{p} = \frac{h}{mv}\), where \(h\) is Planck's constant, \(m\) is mass, and \(v\) is velocity.
Step 2: Record necessary constants and given data. \(h \approx 6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\). The electron's mass, \(m_e \approx 9.11 \times 10^{-31} \, \text{kg}\). The electron's velocity, \(v = 10^7 \, \text{m/s}\).
Step 3: Insert values and compute the wavelength. \[\lambda = \frac{6.626 \times 10^{-34} \, \text{J}\cdot\text{s}}{(9.11 \times 10^{-31} \, \text{kg}) \times (10^7 \, \text{m/s})}\] \[\lambda = \frac{6.626}{9.11} \times 10^{-34 - (-31) - 7} \, \text{m}\] \[\lambda \approx 0.727 \times 10^{-10} \, \text{m}\] \[\lambda \approx 7.27 \times 10^{-11} \, \text{m}\] The result is approximately \(7.3 \times 10^{-11}\) m.