An $\alpha$-particle and a proton are simultaneously accelerated from rest through a potential difference of 0.5Mv, work done by electric field is

$\text{w}={\text{w}}_1+{\text{w}}_{\text{2}}$

${\text{w}}_{\text{1}}={\text{q}}_{\text{1}}\text{V}$                                                          

${\text{w}}_{\text{1}}\text{for}\propto \left(2{\text{He}}^{\text{4}}\right)$                                               

${\text{w}}_{\text{1}}=2\times 1.6\times {10}^{-19}\times 0.5\times {10}^6$                

$=1.6\times {10}^{-13}$                                   

$=1\text{Mev}$

$(1\text{ev}=1.6\times {10}^{-19}\text{J})$

${\text{w}}_{\text{2}}={\text{q}}_{\text{2}}\text{V}$

${\text{w}}_{\text{2}}\text{\hspace{0.17em}for\hspace{0.17em}proton\hspace{0.17em}}\left(1{\text{H}}^{\text{1}}\right)$

${\text{w}}_{\text{2}}=1.6\times {10}^{-19}\times 0.5\times {10}^6$

$=0.5\times 1.6\times {10}^{-13}$

$=0.5\text{\hspace{0.17em}Mev}$

$\therefore \text{w}=\text{1}+\text{0.5}=\text{1.5\hspace{0.17em}Mev}$