91影视

Accelerated potential, $螖V=50,000\text{V}$

The energy of electron in${\text{n}}^{\text{th}}$-orbit:

The energy KE which an object of charge q gains by passing through the potential difference $螖V$is$KE=q螖V$

(a)

If the accelerating potential is $V=50\text{kV}$, then the energy of the accelerated electron would be$E=50\text{keV}$ .

This electron can 'knock out' an electron in the shell with an energy of $E_1=-50\text{keV}$.

By Eq. (1), find the atomic number Z as (note that $n=1$ for the K shell).

$$\begin{gathered} Z=\sqrt{\frac{50路{10}^3\text{eV}}{13.6\text{eV}}} \\ Z鈮�60\backslash \mathrm{hfill} \end{gathered}$$

(b)

If we were to strike an atom with higher atomic number than the one, we've calculated z=60 , we could make a hole but the level must be higher than $n=1$(the K shell).

If n was not larger, then the absolute value of the orbiting electron would be greater (see Eq. (1)) and the value of the accelerating potential would not be great enough.