A \( \beta \)-particle is moving perpendicular to the uniform electric field created by a potential difference of \( 6\times10^{3}\,\mathrm{V} \) across parallel metal plates separated by a vertical distance of 6 cm. What is the acceleration of the beta particle between the plates? (Mass of electron = \( 9.1\times10^{-31}\,\mathrm{kg} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.76\times10^{16}\,\mathrm{m\,s^{-2}} \)
A \( \beta \)-particle is an electron. Between parallel plates the field is uniform, \( E = V/d \), and the electric force \( eE \) gives the acceleration by Newton's second law.
\( \displaystyle d = 6\,\mathrm{cm} = 0.06\,\mathrm{m} \)
\( \displaystyle E = \frac{V}{d} = \frac{6\times10^{3}\,\mathrm{V}}{0.06\,\mathrm{m}} = 10^{5}\,\mathrm{V\,m^{-1}} \)
\( \displaystyle a = \frac{eE}{m} = \frac{1.6\times10^{-19} \times 10^{5}}{9.1\times10^{-31}} \approx 1.76\times10^{16}\,\mathrm{m\,s^{-2}} \)