When the operating potential in an X-ray tube is raised by 1\% , by...

The cutoff wavelength of X-rays produced in an X-ray tube is given by the relation:

$\lambda_{c} = \dfrac{hc}{eV}$

Taking the natural logarithm on both sides, we get:

$\ln \lambda_{c} = \ln \left(\dfrac{hc}{e}\right) - \ln V$

Differentiating the expression for small changes, we obtain:

$\dfrac{\Delta \lambda_{c}}{\lambda_{c}} = - \dfrac{\Delta V}{V}$

Multiplying by $100$ to get the percentage change:

$\dfrac{\Delta \lambda_{c}}{\lambda_{c}} \times 100 = - \left(\dfrac{\Delta V}{V} \times100\right)$

Given that the operating potential is raised by $1\%$, we have $\dfrac{\Delta V}{V} \times100 = 1\%$.

$\dfrac{\Delta \lambda_{c}}{\lambda_{c}} \times100 = -1\%$

The negative sign indicates a decrease. Therefore, the cutoff wavelength decreases by $1\%$.