Question

An $\alpha $ particle and a proton are accelerated from rest by the same potential. Find the ratio of their de-Broglie wavelength.

Answer 1

- Hint: First, we will find out the kinetic energies of both the $\alpha $ and the proton respectively with the formula $KE = \dfrac{1}{2}m{v^2}$. Then we will solve the equations further and find out the equation for momentums of both the particles. Refer to the solution below.

Formula used: $KE = \dfrac{1}{2}m{v^2}$, $p = mv$, $\lambda = \dfrac{h}{p}$.

Complete step-by-step solution:
Let the mass of $\alpha $particle be ${m_\alpha }$.
Let the mass of the proton be ${m_p}$.
Let the velocity of $\alpha $ particle be ${v_\alpha }$.
Let the velocity of the proton be ${v_p}$.
Now, as we know the formula for kinetic energy is $KE = \dfrac{1}{2}m{v^2}$.
Kinetic energy of $\alpha $ particle will be-
$ \Rightarrow K{E_\alpha } = \dfrac{1}{2}{m_\alpha }{v_\alpha }^2$
Multiplying the numerator and denominator by ${m_\alpha }$, we get-
$ \Rightarrow K{E_\alpha } = {\dfrac{{\left( {{m_\alpha }{v_\alpha }} \right)}}{{2{m_\alpha }}}^2}$
As we know that the formula for momentum is $p = mv$. So, from the above equation we get-
$
   \Rightarrow K{E_\alpha } = {\dfrac{{\left( {{p_\alpha }} \right)}}{{2{m_\alpha }}}^2} \\
    \\
   \Rightarrow {p_\alpha }^2 = 2{m_\alpha }{\left( {KE} \right)_\alpha } \\
    \\
   \Rightarrow {p_\alpha } = \sqrt {2{m_\alpha }{{\left( {KE} \right)}_\alpha }} \\
$
Kinetic energy of proton will be-
$ \Rightarrow K{E_p} = \dfrac{1}{2}{m_p}{v_p}^2$
Multiplying the numerator and denominator by ${m_p}$, we get-
$ \Rightarrow K{E_p} = {\dfrac{{\left( {{m_p}{v_p}} \right)}}{{2{m_p}}}^2}$
As we know that the formula for momentum is $p = mv$. So, from the above equation we get-
$
   \Rightarrow K{E_p} = {\dfrac{{\left( {{p_p}} \right)}}{{2{m_p}}}^2} \\
    \\
   \Rightarrow {p_p}^2 = 2{m_p}{\left( {KE} \right)_p} \\
    \\
   \Rightarrow {p_p} = \sqrt {2{m_p}{{\left( {KE} \right)}_p}} \\
$
Now, the work done in accelerating the proton and the $\alpha $ particle will be equal to the kinetic energy acquired. As we know, $W = qV$. Potential difference is the same in both cases. So-
Kinetic energy of $\alpha $ particle in terms of charge and potential difference-
$ \Rightarrow K{E_\alpha } = {q_\alpha }V$
Kinetic energy of $p$ particle in terms of charge and potential difference-
$ \Rightarrow K{E_p} = {q_p}V$
Putting the above values of kinetic energy into the values of momentums, we get-
For $\alpha $ particle-
$
   \Rightarrow {p_\alpha } = \sqrt {2{m_\alpha }{{\left( {KE} \right)}_\alpha }} \\
    \\
   \Rightarrow {p_\alpha } = \sqrt {2{m_\alpha }\left( {{q_\alpha }V} \right)} \\
$
For proton-
$
   \Rightarrow {p_p} = \sqrt {2{m_p}{{\left( {KE} \right)}_p}} \\
    \\
   \Rightarrow {p_p} = \sqrt {2{m_p}\left( {{q_p}V} \right)} \\
$
The formula for de-Broglie wavelength is $\lambda = \dfrac{h}{p}$. Putting the values of momentum from above one by one, we get-
For $\alpha $ particle-
$ \Rightarrow {\lambda _\alpha } = \dfrac{h}{{\sqrt {2{m_\alpha }\left( {{q_\alpha }V} \right)} }}$
For proton-
$ \Rightarrow {\lambda _p} = \dfrac{h}{{\sqrt {2{m_p}\left( {{q_p}V} \right)} }}$
Finding their ratios, we will have-
$
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{\dfrac{h}{{\sqrt {2{m_\alpha }\left( {{q_\alpha }V} \right)} }}}}{{\dfrac{h}{{\sqrt {2{m_p}\left( {{q_p}V} \right)} }}}} \\
    \\
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{h \times \sqrt {2{m_p}\left( {{q_p}V} \right)} }}{{h \times \sqrt {2{m_\alpha }\left( {{q_\alpha }V} \right)} }} \\
    \\
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{\sqrt {{m_p}{q_p}} }}{{\sqrt {{m_\alpha }{q_\alpha }} }} \\
$
As we know that the mass of $\alpha $ particle is 4 times the mass of proton and the charge of $\alpha $ particle is 2 times the charge of proton, we get-
$
   \Rightarrow {m_\alpha } = 4{m_p} \\
    \\
   \Rightarrow {q_\alpha } = 2{q_p} \\
$
Putting the values in the above ratio, we will have-
$
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{\sqrt {{m_p}{q_p}} }}{{\sqrt {{m_\alpha }{q_\alpha }} }} \\
    \\
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{\sqrt {{m_p}{q_p}} }}{{\sqrt {4{m_p}2{q_p}} }} \\
    \\
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \sqrt {\dfrac{1}{4} \times \dfrac{1}{2}} \\
    \\
   \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{1}{{2\sqrt 2 }} \\
$
Thus, the ratio of ${\lambda _\alpha }:{\lambda _p} = 1:2\sqrt 2 $.

Note: It is said that matter has a dual nature of wave-particles. de Broglie waves, named after the pioneer Louis de Broglie, is the property of a material object that differs in time or space while acting like waves. It is likewise called matter-waves. It holds extraordinary likeness to the dual nature of light which acts as particle and wave, which has been demonstrated experimentally.