Dual Nature of Matter

JEE Main level, single correct, medium

26 of 200

Question 26 of 200

A proton and an α\alpha-particle are accelerated from rest by 2V2V and 4V4V potentials, respectively. The ratio of their de-Broglie wavelength is :

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Answer: (C) 4:14:1

The formula to calculate the de Broglie wavelength of the proton is given by
λp=h2mpqpV...(1){\lambda }_{p}=\dfrac{h}{\sqrt{2{m}_{p}{q}_{p}V}}...\left(1\right)
The formula to calculate the de Broglie wavelength of the alpha particle is given by
λα=h2mαqαV′...(2){\lambda }_{\alpha }=\dfrac{h}{\sqrt{2{m}_{\alpha }{q}_{\alpha }V'}}...\left(2\right)
Divide equation (1) by equation (2) and simplify to obtain the required ratio.
λpλα=mαmp×qαV′qpV...(3)\dfrac{{\lambda }_{p}}{{\lambda }_{\alpha }}=\dfrac{\sqrt{{m}_{\alpha }}}{\sqrt{{m}_{p}}}\times \sqrt{\dfrac{{q}_{\alpha }V'}{{q}_{p}V}}...\left(3\right)
Substitute the values of the known parameters into equation (3) to calculate the required ratio.
λpλα=41×2×41×2=4:1\begin{matrix}\dfrac{{\lambda }_{p}}{{\lambda }_{\alpha }} & = & \sqrt{\dfrac{4}{1}}\times \sqrt{\dfrac{2\times 4}{1\times 2}} \\ & = & 4:1\end{matrix}

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