Question

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

Answer 1

Given that,

Short wave length ${\lambda }^{\prime }=\lambda -26pm$

Operating voltage $V^{{}^{\prime }}=1.5V$

We know that,

$\lambda =\frac{hc}{eV}....\left(I\right)$

${\lambda }^{\prime }=\frac{hc}{e{V}^{\prime }}$

Now,

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

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

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

$\lambda -1.5\lambda =-39\times {10}^{-12}$

$0.5\lambda =39\times {10}^{-12}$

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

Now, put the value of $\lambda$ in equation (I)

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

$V=\frac{hc}{e\lambda }$

$V=\frac{6.67\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 78\times {10}^{-12}}$

$V=\frac{20.01\times {10}^5}{124.8}$

$V=16.03\times {10}^{3}volt$

$V=16KV$

Hence, the original value of the operating voltage is $16KV$