Question
The intensity of X-rays from a Coolidge tube is plotted
against wavelength $$\lambda $$ as shown in the figure. The minimum wavelength found is $${\lambda _C}$$ and the wavelength of the $${K_\alpha }$$ line is $${\lambda _K}.$$ As the accelerating voltage is increased
A.
$${\lambda _K} - {\lambda _C}$$ increases
B.
$${\lambda _K} - {\lambda _C}$$ decreases
C.
$${\lambda _K}$$ increases
D.
$${\lambda _K}$$ decreases
Answer :
$${\lambda _K} - {\lambda _C}$$ increases
Solution :
KEY CONCEPT :
In case of Coolidge tube
$${\lambda _{\min }} = \frac{{hc}}{{eV}} = \lambda \left( {{\text{as}}\,{\text{given}}\,{\text{here}}} \right)$$
Thus the cut off wavelength is inversely proportional to accelerating voltage. As $$V$$ increases, $${\lambda _c}$$ decreases. $${\lambda _k}$$ is the wavelength of $${K_ \propto }$$ line which is a characteristic of an atom and does not depend on accelerating voltage of bombarding electron since $${\lambda _k}$$ always refers to a photon wavelength of transition of $${e^ - }$$ from the target element from $$2 \to 1.$$
The above two facts lead to the conclusion that $${\lambda _k} - {\lambda _c}$$ increases as accelerating voltage is increased.