Question

The mass of an electron is 9.1 × 10^–31 kg. If its K.E. is 3.0 × 10^–25 J, calculate its wavelength.

Answer 1

Given

• Mass of electron, \( m = 9.1 \times 10^{-31}\,\text{kg} \).
• Kinetic energy, \( K.E. = 3.0 \times 10^{-25}\,\text{J} \).
• Planck’s constant, \( h = 6.626 \times 10^{-34}\,\text{J s} \).

Step 1: Relation between K.E. and momentum

Kinetic energy of a particle: \[ K.E. = \frac{p^{2}}{2m} \] So, \[ p = \sqrt{2m \cdot K.E.} \]

Step 2: de Broglie wavelength

de Broglie equation: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot K.E.}} \]

Step 3: Substitution

\[ \lambda = \frac{6.626 \times 10^{-34}} {\sqrt{2 \times 9.1 \times 10^{-31} \times 3.0 \times 10^{-25}}} \] \[ \lambda \approx \frac{6.626 \times 10^{-34}} {\sqrt{54.6 \times 10^{-56}}} \] \[ \sqrt{54.6} \approx 7.39,\quad \sqrt{10^{-56}} = 10^{-28} \] \[ \Rightarrow \sqrt{54.6 \times 10^{-56}} \approx 7.39 \times 10^{-28} \] \[ \lambda \approx \frac{6.6 \times 10^{-34}}{7.39 \times 10^{-28}} \approx 0.8967 \times 10^{-6}\,\text{m} = 8.97 \times 10^{-7}\,\text{m} \]

Final Answer

The de Broglie wavelength of the electron is approximately \( \lambda \approx 9.0 \times 10^{-7}\,\text{m} \) (or about \( 9.0 \times 10^{3}\,\text{Å} \)).