Question

An α-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are ${{\lambda }_{\alpha }}\text{ and }{{\lambda }_{p}}$ respectively. The ratio $\dfrac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}$,to the nearest integer, is ___.

Answer 1

We have given the potential difference for alpha particle and proton when it is accelerated from rest and we have to find the ratio of de Broglie wavelength associated with it. So first we will find the de Broglie wavelength of the alpha particle and proton by using the de Broglie formula.
Formula used:

& \lambda =\dfrac{h}{p} \\\ & p=\sqrt{2mK.E.} \\\ & K.E.=eV \\\ \end{aligned}$$ **Complete answer:** The de Broglie wavelength is given as the ratio of the Planck’s constant and the momentum of the particle. Its mathematical representation is $$\lambda =\dfrac{h}{p}\text{ }.............\text{(i)}$$ Now here we have given the voltage difference of alpha particle and the proton and we have to find the de Broglie wavelength. For that we will rewrite the de Broglie wavelength in terms of voltage difference. The kinetic energy of a particle is given as product of charge on the particle and the voltage difference associated with it, i.e. $$K.E.=eV$$ Where e is charge and V is the voltage difference. Now the momentum can be given in term of kinetic energy as $$p=\sqrt{2mK.E.}$$ Substituting value of K.E. in the above equation we get $$p=\sqrt{2meV}$$ Now substituting above value in equation (i) we get, $$\lambda =\dfrac{h}{\sqrt{2meV}}$$ From the above formula we can write the de Broglie wavelength for proton as $${{\lambda }_{p}}=\dfrac{h}{\sqrt{2{{m}_{p}}{{e}_{p}}{{V}_{p}}}}\text{ }................\text{(ii)}$$ Where $${{m}_{p}}$$is the mass of proton, $${{e}_{p}}$$is the charge on proton and $${{V}_{p}}$$is the voltage difference associated with proton. Similarly, for alpha particle de Broglie wavelength is $${{\lambda }_{\alpha }}=\dfrac{h}{\sqrt{2{{m}_{\alpha }}{{e}_{\alpha }}{{V}_{\alpha }}}}\text{ }................\text{(iii)}$$ Where $${{m}_{\alpha }}$$is the mass of the alpha particle, $${{e}_{\alpha }}$$is the charge on the alpha particle and $${{V}_{\alpha }}$$is the potential difference associated with the alpha particle. Now alpha particle has mass four times the mass of proton and charge which is twice the charge of proton, i.e. $$\begin{aligned} & {{m}_{\alpha }}=4{{m}_{p}} \\\ & {{e}_{\alpha }}=2{{e}_{p}} \\\ \end{aligned}$$ Now substituting mass and charge of alpha particle in equation (iii), we get $$\begin{aligned} & {{\lambda }_{\alpha }}=\dfrac{h}{\sqrt{2\left( 4{{m}_{p}} \right)\left( 2{{e}_{p}} \right){{V}_{\alpha }}}} \\\ & \Rightarrow \lambda =\dfrac{h}{\sqrt{16{{m}_{p}}{{e}_{p}}{{V}_{\alpha }}}}\text{ }................\text{(iv)} \\\ \end{aligned}$$ Now using equation (ii) and (iv), is given as $$\begin{aligned} & \dfrac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\dfrac{\dfrac{h}{\sqrt{2{{m}_{p}}{{e}_{p}}{{V}_{p}}}}}{\dfrac{h}{\sqrt{16{{m}_{p}}{{e}_{p}}{{V}_{\alpha }}}}} \\\ & \Rightarrow \dfrac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\dfrac{h}{\sqrt{2{{m}_{p}}{{e}_{p}}{{V}_{p}}}}\times \dfrac{\sqrt{16{{m}_{p}}{{e}_{p}}{{V}_{\alpha }}}}{h} \\\ & \Rightarrow \dfrac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\dfrac{\sqrt{8{{V}_{a}}}}{\sqrt{{{V}_{p}}}} \\\ \end{aligned}$$ Now we have given the potential difference as $${{V}_{a}}={{V}_{p}}=100V$$ substituting this value we get $$\begin{aligned} & \dfrac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\dfrac{\sqrt{8\times 100}}{\sqrt{100}} \\\ & \Rightarrow \dfrac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\sqrt{8}=2 \sqrt{2}\\\ \end{aligned}$$ **Hence the ratio of the de Broglie wavelength of protons to the de Broglie wavelength of alpha particles is 3.** **Note:** The alpha particle is the helium atom which has two protons and it has twice the mass of protons. Also it has two electrons therefore it has twice the charge of proton. Note that de Broglie wavelength shows the wave nature of the light as light has dual nature i.e. of matter and wave.