How are characteristic and cut-off wavelength related to each other for a metal 
Find the binding energy of an L electron in titanium if the wavelength difference between the first line of the K series and its short-wave cut-off is $\Delta \lambda=26 \;\text{pm}$.

Cut-off wavelength including the other x-ray spectrum is the contentious emission of x-rays due to the retardation of electron caused by the target metal.
In the process  some of the electrons are knocked out from the atoms of the target, and the electrons present in the higher energy orbits jumps to the lower energy orbits giving off the characteristic spectrum($K_{\alpha}...,L_{\alpha}...,...$).
As we can see there is no relation between the cut-off wavelength and the characteristic wavelength.
Hence binding energy of L electron can be calculated as follows:

Energy required to $K\to L$ transitions is:
$$E_{K\to L}=\frac{hc}{\lambda}=hcR(Z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$$
$$\implies E_{K\to L}\approx 4509 \;\text{eV}$$
But the binding energy of K electron can be given as:
$$E_{K\to \infty}=\frac{hc}{\lambda}=hcR(Z-1)^2\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]$$
$$\implies E_{K\to \infty}\approx 6012 \;\text{eV}$$
Hence binding energy of L electron can be given as:
$$E_{L\to \infty}=E_{K\to \infty}-E_{K\to L}$$
$$\implies E_{L\to \infty} \approx 1509 \;\text{eV}$$

But my answer comes out to be incorrect and the correct answer is $\approx 500 \;\text{eV}$.
Where as author has provided with a solution which I am unable to comprehend, as I could not understand his logic. Following is his solution:

$$h\nu=\frac{hc}{\lambda}=E_{K}-E_{L}=hcR(Z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$$
$$\implies \lambda=\frac{4}{3hcR(Z-1)^2}$$
Also then energy of the line corresponding to the shortwave cut-off is :
$$E_{K}=\frac{hc}{\lambda-\Delta \lambda}$$
Now
$$E_{L}=E_{K}-\frac{hc}{\lambda}$$
and on substituting the values we get $E_{L}=500 \;\text{eV}$

I don't see any logic in the author's solution. I have no idea what he did. If author is correct please explain what he did and also tell why I am wrong.
 A: Let's first start by defining terms in the expression:
$$
\frac{ h c }{ \lambda } = h \nu = h \ c \ R \left( Z - 1 \right)^{2} \left[ n^{-2} - n'^{-2} \right] \tag{0}
$$
where $h$ is the Planck constant, $c$ is the speed of light in vacuum, $\lambda$($\nu$) is the wavelength(frequency) of a photon necessary to move an electron from energy level $n$ to $n'$, $Z$ is the atomic number of the metal, and $R$ is the modified Ryberg constant for the metal given by:
$$
R = \frac{ R_{o} }{ 1 + \tfrac{ m_{e} }{ M } } \tag{1}
$$
where $m_{e}$ is the rest mass of an electron, $M$ is the mass of the metal nucleus, and $R_{o}$ is the Ryberg constant given by:
$$
R_{o} = \frac{ m_{e} e^{4} }{ 8 \varepsilon_{o}^{2} h^{3} c } \tag{2}
$$
where $e$ is the fundamental charge and $\varepsilon_{o}$ is the permittivity of free space.
Note that we can rearrange Equation 0 in terms of wavelength to get:
$$
\lambda = \frac{ 1 }{ R \left( Z - 1 \right)^{2} } \left( \frac{ n^{2} n'^{2} }{ n'^{2} - n^{2} } \right) \tag{3}
$$
Cutoff
The phenomena of cutoff occurs here because the system is quantized.  That is, a photon incident on a conductor can give energy to an electron.  If the electron is liberated from the conductor, the energy is given by:
$$
E_{\nu} = h \nu = \frac{ h c }{ \lambda } = KE_{max} + \phi \tag{4}
$$
where $KE_{max}$ is the kinetic energy of the electron at the instant it is liberated from the surface of the conductor and $\phi$ is the energy necessary to liberate the electron from the surface of the conductor, also called the work function of the material.  In this expression, the following must be satisfied $KE_{max} \ge 0$.  If we set $KE_{max} = 0$, then we define the cutoff frequency, $\nu_{c}$, or wavelength, $\lambda_{c}$, of the material.  Doing so gives us:
$$
\lambda_{c} = \frac{ h c }{ \phi } \tag{5}
$$
In the problem you present, $\lambda_{c}$ = 26 pm.
Example
In the following I am just plugging in numbers in the expressions provide in your question to illustrate that they will indeed provide the correct result.
Now if we transition from $1$ to $2$, then Equation 3 reduces to:
$$
\lambda = \frac{ 4 }{ 3 R \left( Z - 1 \right)^{2} } \tag{6}
$$
If use the values for titanium, we get a wavelength of $\lambda$ ~ 276 pm.  Then using the following expression:
$$
E_{K} = \frac{ h c }{ \lambda - \lambda_{c} } \tag{7}
$$
we find that $E_{K}$ ~ 4960 eV.
Finally, we find the difference in energy given by:
$$
E_{L} = E_{K} - \frac{ h c }{ \lambda } \tag{8}
$$
The second expression here corresponds to ~4500 eV, thus the difference results in $E_{L}$ ~ 460 eV.
