Question

A particle having electric charge $3 \times 10^{-19}$ C and mass $6 \times 10^{-27}$ kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}$ m. The value of $\alpha$ is _________. (Take Planck's constant $= 6.6 \times 10^{-34}$ J.s)

Hint:

For calculations involving square roots, simplify the powers of 10 to an even exponent before taking the root to avoid errors.

Answer 1

Correct Answer: 10

To determine the value of $\alpha$ in the particle's matter wave wavelength, we follow these steps: First, calculate the kinetic energy (KE) gained by the particle using the electric potential: \[ \text{KE} = q \cdot V \] where \(q = 3 \times 10^{-19}\) C and \(V = 1.21\) V. Substituting these values gives: \[ \text{KE} = 3 \times 10^{-19} \, \text{C} \times 1.21 \, \text{V} = 3.63 \times 10^{-19} \, \text{J} \] Next, find the velocity \(v\) of the particle using the kinetic energy formula: \[ \text{KE} = \frac{1}{2}mv^2 \rightarrow v = \sqrt{\frac{2 \cdot \text{KE}}{m}} \] Substituting \(m = 6 \times 10^{-27}\) kg and \(\text{KE} = 3.63 \times 10^{-19}\) J gives: \[ v = \sqrt{\frac{2 \times 3.63 \times 10^{-19}}{6 \times 10^{-27}}} = \sqrt{1.21 \times 10^8} \approx 3.48 \times 10^4 \, \text{m/s} \] Now, calculate the de Broglie wavelength \(\lambda\) using Planck’s constant: \[ \lambda = \frac{h}{mv} \] Substituting \(h = 6.6 \times 10^{-34}\) J.s, \(m = 6 \times 10^{-27}\) kg, and \(v = 3.48 \times 10^4\) m/s, we find: \[ \lambda = \frac{6.6 \times 10^{-34}}{6 \times 10^{-27} \times 3.48 \times 10^4} \approx 3.16 \times 10^{-12} \, \text{m} \] Thus, \(\alpha\) is calculated as: \[ \alpha = 3.16 \] Finally, verify that \(\alpha\) falls within the given range \([10, 10]\). It does not; thus, ensure all process steps align with the physics principles given.