$\large{X- rays}$

First of all, those who haven't seen, First please see the Mosley's Law In Wikipedia.

From Mosley's Law, we get the Frequency of K-Alpha Line is defined by the equation, f = (2.47)(10^15)((Z-1)^2) Hz, where Z is the Atomic Number of the Target Element.

Now, from this equation, It is quite clear that the Frequency and so the Wavelength of K-Alpha Line doesn't depend on the Voltage or Potential Difference. It just depends on the Atomic Number of the Target Element.

So, in BOTH of the cases, stated in the question, The Wavelength of K-Alpha Line will NOT Change, & that constant wavelength will be defined by the Equation,

(Lambda) = c/f, where C is the speed of Light & f is the above stated frequency. Lets assume this Lamda to be X.

Now, for MOST Energetic X-Rays,

The TOTAL Energy of the Electron has to be converted to the Photon's Energy which clearly states that, (e)(V) = (h)(c/Lambda'), where, e is the charge of Electron, V is the Potential Difference under which The X-Ray Tube is working, h is Planck's Constant & (Lambda') is the Wavelength of Photon emitted or X-Ray.

For first case, Lambda 1 = (hc)/(38080 e)

For second case, Lambda 2 = (hc)/(19040 e)

Therefore, the Difference between the M.E.X-Ray & K-Alpha X-Ray is,

First Case :- (Lambda 1) - X

Second Case :- (Lambda 2) - X

According to the question,

(Lambda 1 - X) = (4)(Lambda 2 - X) ................................................ (1)

Because, the Difference of the Wavelength of Most Energetic X-Ray & K-Alpha X-Ray becomes 4 Times of the difference prior to the change of Voltage.

Now, just Plug in the values of (X),(Lambda 1) & (Lambda 2) from above in the 1 no. Equation & after doing some general calculations, you will get,

Z = 40.93142674 that is Z = 41 (approx.)

So, the atomic number of the target element is 41 which is the answer.