Question

The ratio of de-Broglie wavelengths of proton and $ \alpha $ -particle having same kinetic energy is
(A) $ \sqrt 2 :1 $
(B) $ 2\sqrt 2 :1 $
(C) $ 2:1 $
(D) $ 4:1 $

Answer 1

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}} $ .