Question

In a Thomson set-up for the determination of e/m, electrons accelerated by 2.5 kV enter the region of crossed electric and magnetic fields of strengths $3.6\times {10}^{4}V{M}^{-1}$ and $1.2\times {10}^{-3}T$ respectively and go through undeflected. The measured value of e/m of the electron is equal to

• A. $1.0\times {10}^{11}C/k{g}^{-1}$
• B. $1.76\times {10}^{11}C/k{g}^{-1}$
• C. $1.80\times {10}^{11}C/k{g}^{-1}$
• D. $1.85\times {10}^{11}C/k{g}^{-1}$

Answer 1

The correct option is C $1.80\times {10}^{11}C/k{g}^{-1}$

By using $\frac{e}{m}=\frac{E^2}{2V{B}^2}\Rightarrow \frac{e}{m}=\frac{(3.6\times {10}^4{)}^2}{2\times 2.5\times {10}^3\times (1.2\times {10}^{-3}{)}^2}=1.8\times {10}^{11}C/kg$.