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Q.A proton and an alpha particle have equal momentum. The ratio of their kinetic energies (EpEα)\left(\dfrac{E_p}{E_\alpha}\right) and the ratio of the de Broglie wavelengths associated with them (λpλα)\left(\dfrac{\lambda_p}{\lambda_\alpha}\right) respectively are : (A) 2, 12,\ 1 (B) 1, 21,\ 2 (C) 4, 14,\ 1 (D) 1, 41,\ 4

Sikkim CbseCBSE Class XII Board 2026MCQ· 1mImportance★★★★★
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For equal momentum, kinetic energy is inversely proportional to mass, and de Broglie wavelength is directly proportional to mass. Since the alpha particle has 4 times the mass of a proton, the ratio of kinetic energies is 4:14:1 and the ratio of wavelengths is 1:11:1. The correct option is (C).

The key to this problem lies in two fundamental relationships: the de Broglie wavelength and the connection between kinetic energy and momentum. When two particles have the same momentum, their de Broglie wavelengths become equal — that part is immediate. The kinetic energy, however, depends on mass because Ek=p2/2mE_k = p^2/2m, so the lighter particle has more kinetic energy.

Let’s work through it systematically.

  1. Recall the de Broglie wavelength formula.

    Every moving particle has a wavelength associated with it, given by λ=hp\lambda = \frac{h}{p}, where hh is Planck’s constant and pp is the linear momentum. This is a direct consequence of wave-particle duality — the more momentum a particle has, the shorter its wavelength.

    Since the problem states that the proton and alpha particle have equal momentum, we can write:

pp=pαp_p = p_\alpha

Therefore:

λp=hppandλα=hpα\lambda_p = \frac{h}{p_p} \quad \text{and} \quad \lambda_\alpha = \frac{h}{p_\alpha}

Because pp=pαp_p = p_\alpha, the two wavelengths are identical:

λpλα=1\frac{\lambda_p}{\lambda_\alpha} = 1

  1. Now find the kinetic energy ratio. Kinetic energy is related to momentum by:

Ek=p22mE_k = \frac{p^2}{2m}

This comes from combining Ek=12mv2E_k = \frac12 mv^2 with p=mvp = mv. For equal momentum, the kinetic energy is inversely proportional to mass — a heavier particle moving with the same momentum must be slower, so it has less kinetic energy.

For the proton (mass mpm_p) and alpha particle (mass mαm_\alpha):

EpEα=p2/2mpp2/2mα=mαmp\frac{E_p}{E_\alpha} = \frac{p^2 / 2m_p}{p^2 / 2m_\alpha} = \frac{m_\alpha}{m_p}

  1. Know the masses involved. …

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