When the operating voltage of an X-ray tube is raised to 1.5 times...

The short-wavelength limit of an X-ray tube is given by $\lambda = \dfrac{hc}{eV}$.

Let the initial operating voltage be $V$. The initial short-wavelength limit is $\lambda_1 = \dfrac{hc}{eV}$.

When the voltage is increased to $1.5 V$, the new short-wavelength limit becomes $\lambda_2 = \dfrac{hc}{1.5 eV} = \dfrac{2}{3} \lambda_1$.

The shift in the short-wavelength limit is given as $26 \text{ pm}$. Thus,

$\lambda_1 - \lambda_2 = 26 \text{ pm}$

$\lambda_1 - \dfrac{2}{3} \lambda_1 = 26 \text{ pm}$

$\dfrac{1}{3} \lambda_1 = 26 \text{ pm} \Rightarrow \lambda_1 = 78 \text{ pm}$

Using the value of $hc \approx 1240 \text{ eV nm} = 1240 \times 10^3 \text{ eV pm}$, the initial voltage $V$ is:

$V = \dfrac{hc}{e \lambda_1} = \dfrac{1240 \times 10^3 \text{ eV pm}}{e \times 78 \text{ pm}}$

$V = \dfrac{1240000}{78} \text{ V} \approx 15897 \text{ V} \approx 15.9 \text{ kV}$