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

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 ${\lambda _1}:{\lambda _2} = \sqrt {{m_2}} :\sqrt {{m_1}}$ .