Answer
$v=4.81\times 10^4\frac{m}{s}$
Work Step by Step
We can find the required speed as follows:
$\Delta E=E_2-E_1=(-13.6eV)(\frac{1}{3^2}-\frac{1}{1^2})$
$\Delta E=(12.1eV)(\frac{1.60\times 10^{-19}J}{eV})$
$\Delta E=1.94\times 10^{-18}J$
We know that
$K.E=(\frac{1}{2})mv^2$
$\implies \Delta E=(\frac{1}{2})mv^2$
This simplifies to:
$v=\sqrt{\frac{2\Delta E}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2(1.94\times 10^{-18})}{1.674\times 10^{-27}}}$
$v=4.81\times 10^4\frac{m}{s}$
