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:

• (A) 3 × 10^-4 m
• (B) 2.5 × 10^-8 m
• (C) 2.5 × 10^-6 m
• (D) 1.6 × 10^-8 m

Answer 1

The Correct Option is B

Explanation:
Given:Wavelength of A, λA = 5 × 10^-8 mMomentum of B = 2 × Momentum of AAccording to de-Broglie's equation : λ = $\frac{h}{p}$ ..........(i)where, λ = Wavelengthh = Planck's constantp = MomentumThus, for particle A:${\lambda }_A=\frac{h}{P_A}$ .........(ii)For particle B:${\lambda }_B=\frac{h}{p_B}$ .........(iii)As, ${\mathrm{P}}_{\mathrm{B}}=2\times {\mathrm{p}}_{\mathrm{A}},$ we we have from equation${\lambda }_B=\frac{h}{2p_A}$ ........(iv)Dividing equation (ii) by equation (iv), we get$\frac{{\lambda }_A}{{\lambda }_B}=\frac{\frac{h}{P_A}}{\frac{h}{2P_A}}$$\frac{{\lambda }_A}{{\lambda }_B}=2$${\lambda }_B=\frac{{\lambda }_A}{2}$$=\frac{5\times {10}^{-8}}{2}$$=2.5\times {10}^{-8}$Hence, the correct option is (B).

Answer 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.