Question

In an experiment to determine e/m using Thomson's method, electrons from the cathode accelerate through a potential difference of 1.5 kV. The beam coming out of the anode enters crossed electric and magnetic field of strengths $2\times {10}^{4}V/m$ and $8.6\times {10}^{-4}T$ respectively, The value of e/m. of electron will be

• A. $1.6\times {10}^{11}C/kg$
• B. $1.7\times {10}^{11}C/kg$
• C. $1.8\times {10}^{11}C/kg$
• D. $1.9\times {10}^{11}C/kg$

Answer 1

The correct option is C $1.8\times {10}^{11}C/kg$

Charge on electron $q=1.619\times {10}^{-19}C$

Mass on electron $m=9.1\times {10}^{-31}kg$

e/m ratio is given by: $\frac{1.619\times {10}^{-19}C}{9.1\times {10}^{-31}kg}=1.8\times {10}^{11}C/kg$