9 questions · self-marked practice — reveal the answer and mark yourself.
$\text{E}=\frac{1242}{0.36}=3450\text{eV}(\text{E}_\text{M}-\text{E}_\text{K})$
Energy needed to ionize an organ atom = 16eV
Energy needed to knock out an electron from K-shell
= (3450 + 16)eV = 3466eV = 3.466KeV.
$\text{V}=40\text{Kv}$
$\text{f}=9.7\times10^{18}\text{Hz}$
$\frac{\text{h}}{\text{c}}=\frac{\text{h}}{\text{eV}}$
$\frac{\text{i}}{\text{f}}=\frac{\text{h}}{\text{eV}}$
$\text{h}=\frac{\text{eV}}{\text{f}}\text{V-s}$
$=\frac{\text{eV}}{\text{f}}\text{V-s}=\frac{40\times10^3}{9.7\times10^{18}}$
$=4.12\times10^{-15}\text{eV-s}$
$\lambda=30\text{pm}=30\times10^{-3}\text{nm}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{1242}{30\times10^{-3}}$
$=41.4\times10^3\text{eV}$
Electric field $=\frac{\text{V}}{\text{d}}=\frac{41.4\times10^3}{1.5}$
$=27.6\times10^3\text{V/m}=27.6\text{KV/m}$
| Cell containing vacancy | K | L | M |
| Energy in keV | 69.5 | 11.3 | 2.3 |
$\therefore$ Energy of electrons = ev
This amount of energy ev = energy of L shell The maximum potential difference that can be applied without emitting any electron is 11.3ev.$\Rightarrow\frac{\text{neV}}{\text{t}}=200\Rightarrow\frac{\text{ne}}{\text{t}}\text{V}=200$
$\Rightarrow\text{i}=200/\text{V}=10\text{mA}$
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{\text{hc}}{\text{eV}}=\frac{1242\text{eV}-\text{nm}}{\text{e}\times30\times10^{3}}$
$=414\times10^{-4}\text{nm}=41.4\text{pm}$
Now $\text{E}_\text{K}-\text{E}_\text{L}=\frac{1242}{21.3\times10^{-3}}=58.309\text{Kev}$
$\text{E}_\text{L}=11.3\text{Kev}$
$\text{E}_\text{K}=58.309=11.3=69.609\text{Kev}$
Now, Ve = 69.609KeV
or
V = 69.609KV.