A proton, when accelerated through a potential difference of V = 29...

${\lambda}_{\text{P}} = {\lambda}_{\alpha}$

$or\dfrac{h}{\sqrt{2{m}_{p}{Q}_{P}{V}_{p}}} = \dfrac{h}{\sqrt{2{m}_{\alpha}{Q}_{\alpha}{V}_{\alpha}}}$

$∴ {m}_{\text{p}}{Q}_{\text{p}}{V}_{\text{p}} = {m}_{\alpha}{Q}_{\alpha}{V}_{\alpha}$

$∴ {V}_{\alpha} = \left(\dfrac{{m}_{\text{p}}}{{m}_{\alpha}}\right)\left(\dfrac{{Q}_{\text{p}}}{{Q}_{\alpha}}\right){V}_{p}$

$⇒ {V}_{\alpha} = \left(\dfrac{1}{4}\right)\left(\dfrac{1}{2}\right) \times 29 . 6V = 3 . 7V$