Answer:
7.17 × 10⁷ volts
Explanation:
It is given that,
Voltage to accelerate electrons to hit a copper plate and produce X-rays,
[tex]V_1=39\ kV=39\times 10^3\ V\\[/tex]
Using the conservation of energy for the electrons as :
[tex]:39\times 10^3=\dfrac{1}{2}mv^2v=\sqrt{\dfrac{2V}{m}}[/tex]
m is the mass of electron
[tex]v=\sqrt{\dfrac{2\times 39\times 10^3}{9.1\times 10^{-31}}}\\\\= 2.93 \times 10^1^7m/s[/tex]
[tex]v=2.93\times 10^{17}\ m/s[/tex]
For the proton, mass,
[tex]m'=1.67\times 10^{-27}\ kg[/tex]
Now using the conservation of energy for the protons. We get :
[tex]V=E=\dfrac{1}{2}m'v^2V=\dfrac{1}{2}\times 1.67\times 10^{-27}\times (2.93\times 10^{17})^2V=7.17\times 10^7\ volts[/tex]
