Question

If $10,000V$ are applied across an X-ray tube, find the ratio of wavelength of the incident electrons and the shortest wavelength of X-ray coming out of the X-ray tube, given $e/m$ of electron $=1.8\times {10}^{11}Ck{g}^{-1}$.

• A. $1:10$
• B. $10:1$
• C. $5:1$
• D. $1:5$

Answer 1

Answer: (A)

$1:10$

$V=10kV={10}^{4}volts$

The shortest wavelength is given as, ${\lambda }_1=hc/e{V}_o$

The de-broglie wavelength is given as ${\lambda }_2=h/p=h/\sqrt{2\times e{V}_o\times m_e}$ wher $m_e$ is the mass of the electron, $p$ is the momentum.

We can further write de-broglie wavelength as

${\lambda }_2=hc/\left(e{V}_o\sqrt{2V_o{c}^2/\frac{e}{m_e}}\right)=hc/10e{V}_o$

$\Rightarrow {\lambda }_2={\lambda }_1/10$

So ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced is ${\lambda }_2:{\lambda }_1=1:10$