Answer
D
Work Step by Step
We know that
$W_{\circ}=\frac{K_2-K_1\frac{\lambda_1}{\lambda_2}}{\frac{\lambda_1}{\lambda_2}-1}$
We plug in the known values to obtain:
$W_{\circ}=\frac{2.57eV-(0.550eV)(\frac{433.9nm}{253.5nm})}{(\frac{433.9nm}
{253.5nm})-1}$
$W_{\circ}=2.29eV$
Now $h=\frac{(W_{\circ}+K_1)\lambda_1}{c}$
We plug in the known values to obtain:
$h=\frac{(2.29eV+0.550eV)(433.9nm)}{3\times 10^8m/s}$
$h=656.9\times 10^{-36}J.s$
$\implies h=6.57\times 10^{-34}J.s$
Thus, the correct option is (D).