Question

If a particle is moving with a momentum of (2 × 10¹⁰) h kg·m/s, then the de Broglie wavelength associated with it (in angstrom) is (where h is Planck's constant)

• A. 1.5
• B. 2.5
• C. 1.0
• D. 0.5
• E. 0.75

Hint:

The de Broglie wavelength is inversely proportional to the momentum of the particle. When the momentum is expressed as a multiple of Planck's constant, the calculation simplifies significantly.

Answer 1

The Correct Option is D

The de Broglie wavelength λ of a particle is given by the formula:

λ = h / p

where:

• h is Planck's constant (h = 6.626 × 10^-34 J·s),
• p is the momentum of the particle.

We are given that the momentum p = (2 × 10¹⁰) h. Substituting this into the de Broglie wavelength formula:

λ = h / (2 × 10¹⁰ h)

Simplify the expression:

λ = 1 / (2 × 10¹⁰)

Now, calculate the value:

λ = 5 × 10^-11 m

Since the question asks for the wavelength in angstroms (1 Å = 10^-10 m), convert the wavelength from meters to angstroms:

λ = 5 × 10^-11 m × (1 Å / 10^-10 m) = 0.5 Å

Thus, the de Broglie wavelength associated with the particle is 0.5 Å. Hence, the correct answer is (D) 0.5.