What will be the ratio of de-Broglie wavelengths of proton and a particle of same kinetic energy?

Answer

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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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15 Questions 15 Marks 12 Mins

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.