Question

The short-wavelength limit shifts by $26\text{pm}$ when the operating voltage in an X-ray tube is increased to $1.5$ times the original value. What was the original value of the operating voltage?

• A. $25.56\text{kV}$
• B. $36.78\text{kV}$
• C. $15.93\text{kV}$
• D. $8.12\text{kV}$

Answer 1

The correct option is C $15.93\text{kV}$

At, operating voltage $=V$, wavelength $=\lambda$

And at operating voltage, $V^{\prime }=1.5V$,
wavelength $={\lambda }^{{}^{\prime }}=\lambda -26\text{pm}$

As, $\lambda =\frac{hc}{eV}$

$\Rightarrow \lambda V={\lambda }^{{}^{\prime }}{V}^{{}^{\prime }}$

$\Rightarrow \lambda V=(\lambda -26\times {10}^{-12})\times 1.5V$

$\Rightarrow \lambda =1.5\lambda -1.5\times 26\times {10}^{-12}$

$\therefore \lambda =78\times {10}^{-12}\text{m}$

$\Rightarrow V=\frac{hc}{e\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 78\times {10}^{-12}}$

$=15.93\times {10}^3\text{V}=15.93\text{kV}$

Hence, $\left(C\right)$ is the correct answer.

Why this question? Key Concept: Cut-off wavelength for a material depends on the applied voltage only.