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: Recall the de-Broglie wavelength formula. The de-Broglie wavelength (\(\lambda\)) of a particle is given by: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] where \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(v\) is its velocity.
Step 2: List the required constants and given values. - Planck's constant, \(h \approx 6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\). - Mass of an electron, \(m_e \approx 9.11 \times 10^{-31} \, \text{kg}\). - Velocity of the electron, \(v = 10^7 \, \text{m/s}\).
Step 3: Substitute the values and calculate the wavelength. \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J}\cdot\text{s}}{(9.11 \times 10^{-31} \, \text{kg}) \times (10^7 \, \text{m/s})} \] \[ \lambda = \frac{6.626}{9.11} \times 10^{-34 - (-31) - 7} \, \text{m} \] \[ \lambda \approx 0.727 \times 10^{-10} \, \text{m} \] \[ \lambda \approx 7.27 \times 10^{-11} \, \text{m} \] This value is approximately \(7.3 \times 10^{-11}\) m.