Respuesta :
Thomson experiment he calculated the charge to mass ratio just be passing the fundamental charge through a tube
He calculated the charge to mass ratio just by finding the deflection of charge while it is passing through the constant electric field
so here we will use the deflection as following
let say it passes the field of length "L"
so here we have
[tex]t = \frac{L}{v}[/tex]
now in the same time if it deflect by some distance
[tex]\delta y = \frac{1}{2}at^2[/tex]
[tex]\delta y = \frac{1}{2}{eE}{m}t^2[/tex]
now by solving this equation we can find e/m ratio
so here correct answer will be
the electron's charge-to-mass ratio
Answer:
Charge-to-mass ratio
Explanation:
Thomson's experiment showed that the cathode rays are made up of negatively charged particles. According to Thomson's experiment, he showed an expression for electron's charge to mass ratio as :
[tex]\dfrac{e}{m}=\dfrac{E^2}{2B^2V}[/tex]
where
e is charge on electron
m is the electron's mass
E is electric field
B is magnetic field
V is potential difference
And the ratio of [tex]\dfrac{e}{m}=1.76\times 10^{11}\ C/kg[/tex]
Hence, the correct option is (d) "charge-to-mass ratio".
