Question:

Two particles A and B are in motion and the wavelength of A is 5 × 10-8 m. The wavelength of B, so that its momentum is double that of A, will be:

Updated On: Jun 26, 2024
  • (A) 3 × 10-4 m
  • (B) 2.5 × 10-8 m
  • (C) 2.5 × 10-6 m
  • (D) 1.6 × 10-8 m
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The Correct Option is B

Approach Solution - 1

Explanation:
Given:Wavelength of A, λA = 5 × 10-8 mMomentum of B = 2 × Momentum of AAccording to de-Broglie's equation : λ = hp ..........(i)where, λ = Wavelengthh = Planck's constantp = MomentumThus, for particle A:λA=hPA .........(ii)For particle B:λB=hpB .........(iii)As, PB=2×pA, we we have from equationλB=h2pA ........(iv)Dividing equation (ii) by equation (iv), we getλAλB=hPAh2PAλAλB=2λB=λA2=5×1082=2.5×108Hence, the correct option is (B).
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Approach Solution -2

According to quantum theory, the energy associated with a photon is given by

E = hf   …(i)

Where

  • h is Planck’s constant
  • f is the frequency

According to the mass-energy equivalence principle, the energy of a photon is

E = mc2    …(ii)

Where

  • m is the mass
  • c is the speed of light

From equation (i) and (ii), we have

hf = mc2

But frequency, f = c/λ

Where λ is wavelength

⇒ hc/λ = mc2

⇒ λ = h/mc

Instead of photon, we have material particle of mass m moving with velocity v, then

λ = h/mv

Where mv = p, momentum of the particle. Therefore

λ = h/p

Above expression is known as the expression for a de-Broglie wavelength that shows the wavelength associated with a particle of mass m moving with velocity v.
 

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