Answer
See the detailed answer below.
Work Step by Step
a) Since we know that the potential energy is given by
$$PE=\frac{1}{2}kx^2$$
and since $k$ is constant, the magnitude of the potential energy had to be positive all the time.
Solving for $k$;
$$k=\dfrac{2PE}{ x^2}$$
and in this case, $x=r-r_0$;
$$k=\dfrac{2PE}{ (r-r_0)^2}$$
We can use the same given graph of 29-17, but we need to shift the graph (the two curves) upward only by 4.5 eV so that the potential energy at the lowest point is now zero.
Plugging the data from the figure into the previous formula;
$$k=\dfrac{2\times 4.5 \times1.6\times10^{-19}}{ ([0.13\times10^{-9}]-[0.074\times10^{-9}] )^2}\approx \color{red}{460} \rm \;N/m$$
Noting that this answer might not be accurate since the data on the graph is not precisely accurate as well. So you can get a different result and it is okay.
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b) We know that the wavelength is given y
$$\lambda=\dfrac{c}{f}$$
$$\lambda=\dfrac{c}{\frac{1}{2\pi }\sqrt{\dfrac{k}{m}}}$$
The author told us to use only half the mass of the hydrogen atom;
$$\lambda=\dfrac{c}{\frac{1}{2\pi }\sqrt{\dfrac{2k}{m_H}}}$$
$$\lambda=2\pi c \sqrt{\dfrac{m_H}{2k}}$$
Plugging the known;
$$\lambda=2\pi \times3\times10^8\sqrt{\dfrac{1.67\times10^{-27}}{2\times 480}}= \color{red}{2.486\times 10^{-6}}\;\rm m$$
