Question

In Thomsons method, a beam of electrons accelerated through a p.d. of $285V$, passes undeflected through perpendicular electric and magnetic fields of intensities ${10}^{5}V/m$ and ${10}^{-2}Wb/m^2$ respectively. Then the value of e/m of electron is:

• A. $1.75\times {10}^{11}C/kg$
• B. $1.66\times {10}^{11}C/kg$
• C. $1.84\times {10}^{11}C/kg$
• D. $1.89\times {10}^{11}C/kg$

Answer 1

The correct option is A $1.75\times {10}^{11}C/kg$

Now, if the beam of electrons passes undeflected
$VB=E$
$V={10}^{7}m/s$
and as we know from Thomson experiment that if the beam of electron passes undeflected then their c/m ratio is $1.758\times {10}^{11}C/kg$
As, velocity $={10}^{7}m/s=\sqrt{\frac{2eV}{m}}$
$\Rightarrow \frac{10/4}{2\times 285}\frac{e}{m}\Rightarrow 1.75\times {10}^{11}\approx \frac{e}{m}$