Question

If an X-ray tube operates at the voltage of $10kV$, find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of electron is $1.8\times {10}^{11}Ckg$.

• A. $1$
• B. $0.1$
• C. $1.8$
• D. $1.2$

Answer 1

The correct option is B $0.1$

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

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 $0.1$