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Dual Nature of Radiation question

2017 · Shift 0 · Q69
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Dual Nature of Radiation question

2017 · Shift 0 · Q69

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA{\lambda _A}λA​ to λB{\lambda _B}λB​ after the collision is:
  1. A
    λAλB=13{{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}λB​λA​​=31​
  2. B
    λAλB=2{{{\lambda _A}} \over {{\lambda _B}}} = 2λB​λA​​=2
  3. C
    λAλB=23{{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}λB​λA​​=32​
  4. D
    λAλB=12{{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}λB​λA​​=21​
View written solutionFree

Correct answer: B

  1. Use 1D elastic collision formulas

Let particle AAA have mass m1=mm_1=mm1​=m and initial velocity u1=vu_1=vu1​=v. Let particle BBB have mass m2=m2m_2=\dfrac{m}{2}m2​=2m​ and initial velocity u2=0u_2=0u2​=0.

For a head-on elastic collision:

v1=m1−m2m1+m2u1+2m2m1+m2u2v_1=\frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2v1​=m1​+m2​m1​−m2​​u1​+m1​+m2​2m2​​u2​

v2=2m1m1+m2u1+m2−m1m1+m2u2v_2=\frac{2m_1}{m_1+m_2}u_1 + \frac{m_2-m_1}{m_1+m_2}u_2v2​=m1​+m2​2m1​​u1​+m1​+m2​m2​−m1​​u2​

Since u2=0u_2=0u2​=0,

=\frac{\frac m2}{\frac{3m}{2}}v =\frac{v}{3}$$ $$v_B'=\frac{2m}{m+\frac m2}v =\frac{2m}{\frac{3m}{2}}v =\frac{4v}{3}$$ 2. **Find momenta after collision** De-Broglie wavelength is $$\lambda=\frac{h}{p}$$ So we first compute momenta: For particle $A$: $$p_A'=m\left(\frac v3\right)=\frac{mv}{3}$$ For particle $B$: $$p_B'=\frac m2\left(\frac{4v}{3}\right)=\frac{2mv}{3}$$ 3. **Take ratio of de-Broglie wavelengths** Since $\lambda \propto \dfrac{1}{p}$, $$\frac{\lambda_A}{\lambda_B}=\frac{h/p_A'}{h/p_B'}=\frac{p_B'}{p_A'}$$ Thus, $$\frac{\lambda_A}{\lambda_B}=\frac{\frac{2mv}{3}}{\frac{mv}{3}}=2$$ 4. **Match with options** $$\frac{\lambda_A}{\lambda_B}=2$$ So the correct option is **B**.
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