Question

$\Delta \lambda$ is the difference between the wavelength of $K_{\alpha }$ line and the minimum wavelength of the continuous X-ray spectrum when the X-ray tube is operated at a voltage V. If the operating voltage is changed to $\frac{V}{3}$, then the above difference is $\Delta {\lambda }^{\prime }$. Then

• A. $\Delta {\lambda }^{{}^{\prime }}=5\Delta \lambda$
• B. $\Delta {\lambda }^{{}^{\prime }}=4\Delta \lambda$
• C. $\Delta {\lambda }^{{}^{\prime }}=3\Delta \lambda$
• D. $\Delta {\lambda }^{{}^{\prime }}<3\Delta \lambda$

Answer 1

Answer: (D)

$\Delta {\lambda }^{{}^{\prime }}<3\Delta \lambda$

$wavelengthof{K}_{\alpha }line$

$\lambda =\frac{hc}{E_K-E_L}$

$\Delta \lambda =\frac{hc}{E_K-E_L}-\frac{hc}{eV}$

$\Delta {\lambda }^{{}^{\prime }}=\frac{hc}{E_K-E_L}-\frac{3hc}{eV}$

$\therefore \Delta {\lambda }^{{}^{\prime }}<3\Delta \lambda$

So, the answer is option (D).