Question

The de Broglie wavelength of an electron with kinetic energy of \( 2.5 \) eV is (in m): \[ (1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}, \quad m_e = 9 \times 10^{-31} \text{ kg}) \]

• A. \( \frac{h \times 10^{-25}}{\sqrt{72}} \)
• B. \( \frac{h \times 10^{25}}{\sqrt{72}} \)
• C. \( \frac{\sqrt{72}}{h \times 10^{-25}} \)
• D. \( \frac{\sqrt{72}}{h \times 10^{25}} \)

Hint:

The de Broglie wavelength is inversely proportional to momentum. For non-relativistic electrons, use \( \lambda = \frac{h}{\sqrt{2m_eK}} \).

Answer 1

The Correct Option is B

Step 1: Applying the de Broglie Wavelength Formula 
The de Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] where \( p \) is the momentum of the electron. The kinetic energy \( K \) is related to momentum by: \[ p = \sqrt{2 m_e K} \] Substituting this into the de Broglie equation: \[ \lambda = \frac{h}{\sqrt{2 m_e K}} \] 

Step 2: Substituting Given Values 
Given that: \[ K = 2.5 \text{ eV} = 2.5 \times 1.6 \times 10^{-19} \text{ J} = 4.0 \times 10^{-19} \text{ J} \] \[ m_e = 9 \times 10^{-31} \text{ kg} \] \(h = 6.626 \times 10^{-34} \text{ Js}\) We substitute these values: \[ \lambda = \frac{h}{\sqrt{2 (9 \times 10^{-31}) (4.0 \times 10^{-19})}} \] \[ = \frac{h}{\sqrt{72} \times 10^{-25}} \] Rewriting: \[ \lambda = \frac{h \times 10^{25}}{\sqrt{72}} \] 

Final Answer: The correct choice is Option (2): \( \frac{h \times 10^{25}}{\sqrt{72}} \).