The potential energy of a system increases, if work is done

Explanation

`r=0.53Å =0.53 xx 10^(-10)m " Charge of electron =-e Charge of proton =e"`
Potential energy `=(1)/(4pi epsi_(0)) .(e(-e))/(1)=(-e^(2))/(4pi epsi_(0)r)`
`V=(-(1.6)^(2) xx 10^(-38) xx 9 xx 10^(9))/(0.53 xx 10^(-10))=(-(1.6)^(2) xx 10^(-38) xx 9 xx 10^(9))/(0.53 xx 10^(-10) xx 1.6 xx 10^(-19))eV =(-1.6 xx 9)/(0.53)eV =-2.7eV`
Total energy `=-27.2eV+13.6eV =-13.6eV`
The work required =13.6eV
At 1.06A distance, PE=-13.6eV
It is brought to zero by adding +13.6eV
PE at 0.53 is -13.6eV and energy needed is 13.6eV