When 40 kV is applied across an X-ray tube, X-rays are obtained...

The maximum energy of an X-ray photon is equal to the kinetic energy of the incident electron.

E_{\text\max} = eV

Given $V = 40 \text{ kV}$, the energy in electron-volts is:

E_{\text\max} = 40 \times 10^3 \text{ eV}

Using the relation E_{\text\max} = h \nu_{\text\max}, the Planck constant $h$ is given by:

h = \dfrac{E_{\text\max}}{\nu_{\text\max}}

Substituting the given values:

$h = \dfrac{40 \times 10^3}{9.7 \times 10^{18}} \text{ eV s}$

$h \approx 4.12 \times 10^{-15} \text{ eV s}$