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 $3.6 \times 10 ^ { 4 } \mathrm { Vm } ^ { - 1 }$ 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-kg^-1
• B. $1.76 \times 10 ^ { 11 }$ C-kg^-1
• C. $1.80 \times 10 ^ { 11 }$ C-kg^-1
• D. $1.85 \times 10 ^ { 11 }$ C-kg^-1

Answer 1

Answer: (C)

$1.80 \times 10 ^ { 11 }$ C-kg^-1

Answer 2

By using $\frac { e } { m } = \frac { E ^ { 2 } } { 2 V B ^ { 2 } }$

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