Question

In producing X rays, a beam of electrons accelerated by a potential difference V is made to strike a metal target. The value of the V, when X-rays will have the lowest wavelength of $0.3094A^0$, is

• A. $10kV$
• B. $20kV$
• C. $30kV$
• D. $40kV$

Answer 1

The correct option is D $40kV$

$\lambda =0.3094A^{\circ }=0.03094nm$
$energy=\frac{hc}{\lambda }$

$=\frac{1240eV-nm}{0.03094nm}(hc=1240eV-nm)$

$=40000eV$
$energy=p.d.\times e$
$40000eV=p.d.\times e$
$p.d.=40000V$
$=40\times {10}^{3}V$
$=40KV.$

So, the answer is option (D).