A proton and an $α$ -particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an $α$ -particle is

$\sqrt{2} : 1$

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$2 : 1$

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$2 \sqrt{2} : 1$

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$\frac{1}{2 \sqrt{2}} : 1$

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Solution

The correct option is C
$2 \sqrt{2} : 1$

$λ = \frac{h}{\sqrt{2 m Q V}} \Rightarrow λ \propto \frac{1}{\sqrt{m Q}} \Rightarrow \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α} Q_{α}}{m_{p} Q_{p}}} = \sqrt{\frac{4 m_{p} \times 2 Q_{p}}{m_{p} \times Q_{p}}} = 2 \sqrt{2}$

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A proton and an α -particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an α-particle is

The correct option is C 2√2:1 λ=h√2mQV⇒λ∝1√mQ⇒λpλα=√mαQαmpQp=√4mp×2Qpmp×Qp=2√2

A proton and an $α$ -particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an $α$ -particle is

The correct option is C
$2 \sqrt{2} : 1$

$λ = \frac{h}{\sqrt{2 m Q V}} \Rightarrow λ \propto \frac{1}{\sqrt{m Q}} \Rightarrow \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α} Q_{α}}{m_{p} Q_{p}}} = \sqrt{\frac{4 m_{p} \times 2 Q_{p}}{m_{p} \times Q_{p}}} = 2 \sqrt{2}$