Question

If the kinetic energy of a moving particle is \( E \), then the de-Broglie wavelength is

• A. \( \lambda = h \sqrt{2mE} \)
• B. \( \lambda = \frac{h}{\sqrt{2mE}} \)
• C. \( \lambda = \frac{h}{\sqrt{2E}} \)
• D. \( \lambda = \frac{hE}{\sqrt{2mE}} \)

Hint:

The de-Broglie wavelength is given by \( \lambda = \frac{h}{\sqrt{2mE}} \), where \( h \) is Planck's constant, \( m \) is mass, and \( E \) is kinetic energy.

Answer 1

The Correct Option is B

Step 1: Use de-Broglie wavelength formula.
The de-Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} \] where \( p = \sqrt{2mE} \) is the momentum. Thus: \[ \lambda = \frac{h}{\sqrt{2mE}} \]
Final Answer: \[ \boxed{\frac{h}{\sqrt{2mE}}} \]