Answer
$\lambda=6.10\times10^{-11}m$.
Work Step by Step
This is identical to a problem done in the book (Example 28–7, page 818). We use Z = 42 for molybdenum, a starting energy level of n = 3, and a final energy level of n’ = 1.
$$\frac{1}{\lambda}=(1.097\times10^7m^{-1})(42-1)^2(\frac{1}{1^2}-\frac{1}{3^2})=1.639\times10^{10}m^{-1}$$
$$\lambda=6.10\times10^{-11}m=0.061\;nm$$
We do not expect perfect agreement with the measured value of 0.063 nm. This situation is not the same as a hydrogen atom with Z=41, because there is partial shielding due to electrons in the n = 2 shell. In effect, the effective nuclear charge is even smaller than 41. As we see from our calculation, this would lead to a larger wavelength, which would be closer to the observed value of 0.063 nm.
