A proton and an $α - p a r t i c l e$ accelerated through same voltage.The ratio of their de-broglie wavelength will be :

1 : 2

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2 \sqrt{2} : 1

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\sqrt{2} : 1

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2 : 1

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Solution

The correct option is A $1 : 2$
$λ_{α} = λ_{P}$
$\frac{h}{\sqrt{2 m_{α} q_{α} V}} = \frac{h}{\sqrt{2 m_{P} q_{P} V}}$

and we know
$m_{α} = 4 m_{P}$
$2 q_{α} = q_{P}$

$\frac{λ_{P}}{λ_{a}} = \frac{1}{2}$

Hence (A) option is correct

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A proton and anα−particle accelerated through same voltage.The ratio of their de-broglie wavelength will be :

The correct option is A 1:2λα=λPh√2mαqαV=h√2mPqPVand we know mα=4mP2qα=qPλPλa=12Hence (A) option is correct

A proton and an $α - p a r t i c l e$ accelerated through same voltage.The ratio of their de-broglie wavelength will be :

The correct option is A $1 : 2$
$λ_{α} = λ_{P}$
$\frac{h}{\sqrt{2 m_{α} q_{α} V}} = \frac{h}{\sqrt{2 m_{P} q_{P} V}}$

and we know
$m_{α} = 4 m_{P}$
$2 q_{α} = q_{P}$

$\frac{λ_{P}}{λ_{a}} = \frac{1}{2}$

Hence (A) option is correct