Answer
The new wavelength is $~~382~nm$
Work Step by Step
We can write an expression for the photoelectric effect:
$hf = K_{max}+\Phi$
$\frac{hc}{\lambda} = K_{max}+\Phi$
Then:
$\frac{hc}{\lambda_1} = K_{max,1}+\Phi$
$\frac{hc}{\lambda_2} = K_{max,2}+\Phi$
We can subtract the first equation from the second equation:
$\frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = K_{max,2} - K_{max,1}$
$hc~(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}) = 1.43~eV - 0.710~eV$
$\frac{1}{\lambda_2} - \frac{1}{\lambda_1} = \frac{0.720~eV}{hc}$
$\frac{1}{\lambda_2} = \frac{1}{\lambda_1} + \frac{0.720~eV}{hc}$
$\frac{1}{\lambda_2} = \frac{1}{491\times 10^{-9}~m} + \frac{(0.720~eV)(1.6\times 10^{-19}~J/eV)}{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}$
$\frac{1}{\lambda_2} = 0.0026158\times 10^9~m^{-1}$
$\lambda_2 = 382~nm$
The new wavelength is $~~382~nm$
