Answer Hint: It is known that the de-Broglie wavelength of a particle is given by, $ \lambda = \dfrac{h}{p} $ where is the Planck’s constant and $ p $ is the momentum of the particle. The value of Planck’s constant is given by, $ h = 6.626 \times {10^ - }^{34}Js $ . Complete step by step answer: Here, we have given two particles one is a proton and another is an alpha particle with the same kinetic energy. Now, we know that the de-Broglie wavelength of a particle is given by, $ \lambda = \dfrac{h}{p} $ where is the Planck’s constant and $ p $ is the momentum of the particle. The value of Planck’s constant is given by, $ h = 6.626 \times {10^ - }^{34}Js $ .So, the de-Broglie wavelength of the alpha particle will be, $ {\lambda _\alpha } = \dfrac{h}{{{p_1}}} $ where, $ {p_1} $ is the momentum of the alpha particle.The de-Broglie wavelength of the proton will be $ {\lambda _p} = \dfrac{h}{{{p_2}}} $ , $ {p_2} $ is the momentum of the proton.Now, we have given here the kinetic energy of both the particles are same and kinetic energy of a particle in terms of momentum can be written as, $ E = \dfrac{{{p^2}}}{{2m}} $ Or, momentum can be written as $ p = \sqrt {2mE} $ So, we can write, the kinetic energy of the alpha particle $ {E_\alpha } = \dfrac{{{p_1}^2}}{{2{m_\alpha }}} $ And the kinetic energy of the proton is $ {E_p} = \dfrac{{{p_2}^2}}{{2{m_p}}} $ .Now, we have here the kinetic energies are equal. So we can write, $ {E_p} = {E_\alpha } $ Also we know that mass of an alpha particle is four times the proton so we can write, $ {m_\alpha } = 4{m_p} $ Putting these values in de-Broglie’s wavelength and dividing we have, $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \dfrac{{\dfrac{h}{{{p_1}}}}}{{\dfrac{h}{{{p_2}}}}} = \dfrac{{{p_1}}}{{{p_2}}} $ Or, $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \dfrac{{\sqrt {2{m_\alpha }E} }}{{\sqrt {2{m_p}E} }} $ Putting, $ {m_\alpha } = 4{m_p} $ we get, Or, $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \dfrac{{\sqrt {2 \cdot 4{m_p}E} }}{{\sqrt {2{m_p}E} }} $ Or, $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = 2\dfrac{{\sqrt {2{m_p}E} }}{{\sqrt {2{m_p}E} }} $ Or, $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = 2 $ So, the ratio of their de-Broglie wavelength is, $ 2:1 $Hence, option (C ) is correct. Note: We can observe that if the kinetic energy of the particles are equal then the ratio of their de-Broglie wavelength depends on the square root of their masses. In general we can write the ratio of the de-Broglie’s wavelength of two particles is $ {\lambda _1}:{\lambda _2} = \sqrt {{m_2}} :\sqrt {{m_1}} $ .
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De-Broglie wavelength: The wavelength of any charged particle due to its motion is called the de-Broglie wavelength. The de Broglie wavelength of electrons can be calculated from Planks constant h divided by the momentum of the particle. λ = h/p where λ is de Broglie wavelength, h is Plank's const, and p is the momentum. In terms of Energy, de Broglie wavelength of charge particle: \(λ=\frac{h}{\sqrt{2mE}}\) where λ is de Broglie wavelength, h is Plank's const, m is the mass of the particle, and E is the energy of the particle. Calculation: mp and qp are the mass and charge of the proton respectively mα and qα is the mass and charge of the alpha particle respectively. Energy (E) is the same for both: 4mp = mα and 2qp = qα Since \(λ=\frac{h}{\sqrt{2mE}}\) \(\frac{λ_p}{\lambda_\alpha }=\sqrt {\frac{m_\alpha }{m_p}} =\sqrt (4m_p/m_p) = 2\) Thus ratio = 2 :1 So, the correct answer is option 4. India’s #1 Learning Platform Start Complete Exam Preparation
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