Question

Find the minimum wavelength of X-ray produced if 10 kV potential difference is applied across the anode and cathode of the tube.

• A. 12.4 A${}^{\circ }$
• B. 12.4 nm
• C. 1.24 nm
• D. 1.24 A${}^{\circ }$

Answer 1

The correct option is A 12.4 A${}^{\circ }$

If electrons are accelerated to a velocity $v$ by a potential difference V and then allowed to collide with a metal target, the maximum frequency of the X-rays emitted is given by the equation:
$m{v}^2=eV=h\nu$

or, ${\lambda }_{min}=\frac{1240}{{10}^3}\ast {10}^{-9}=124nm=12.4A^{\circ }$

So, the answer is option (D).