Question

An electron beam has an aperture \(2\,\text{mm}^2\). A total of \(6.0\times10^{15}\) electrons pass through any perpendicular cross-sectional area per second. The current density of the beam is

• A. \(19.2\times10^{-10}\,\text{A m}^{-2}\)
• B. \(9.6\times10^{-4}\,\text{A m}^{-2}\)
• C. \(9.6\times10^{2}\,\text{A m}^{-2}\)
• D. \(4.8\times10^{2}\,\text{A m}^{-2}\)

Hint:

For electron beam problems: \[ I=ne \] \[ J=\frac{I}{A} \] Always convert \(\text{mm}^2\) into \(\text{m}^2\): \[ 1\,\text{mm}^2=10^{-6}\,\text{m}^2 \]

Answer 1

The Correct Option is D

Concept: Current density is defined as \[ J=\frac{I}{A} \] where \[ I=\text{current} \] and \[ A=\text{cross-sectional area} \] Current due to moving electrons is \[ I=ne \] where \[ n=\text{number of electrons passing per second} \] and \[ e=1.6\times10^{-19}\,\text{C} \]

Step 1:Calculate the current. Given, \[ n=6.0\times10^{15}\ \text{s}^{-1} \] Therefore, \[ I=ne \] \[ I=(6.0\times10^{15})(1.6\times10^{-19}) \] \[ I=9.6\times10^{-4}\,\text{A} \]

Step 2: Convert the area into SI units. \[ A=2\,\text{mm}^2 \] \[ A=2\times10^{-6}\,\text{m}^2 \]

Step 3: Calculate current density. \[ J=\frac{I}{A} \] \[ J=\frac{9.6\times10^{-4}}{2\times10^{-6}} \] \[ J=4.8\times10^{2}\,\text{A m}^{-2} \]

Step 4: State the answer. \[ { J=4.8\times10^{2}\,\text{A m}^{-2} } \] Hence, the correct option is \[ {(D)} \]