Question

Derive the equation e/m = (3.29x10⁶)*(Va)/(I²)(R²) for the charge to mass ratio of an electron using the following equations:

(1): (e)(Va) = 1/2 mv²

(2): R = mv/eB

(3): B = ((4/5)^3/2 ) * (µ₀)(N)(I)/r where µ₀ = 4π × 10⁻⁷ H/m ; N = 130; and r = 0.15m

Answer 1

Electron are one of most energetic particle of the atom . When electrons are thermally emitted from the surface that the produce a potential of voltage (Va) as stated in the question. With the help of the quantities like the energy of the electron and the quantities like the accelerated energy and the gain the charge to mass ratio of the electron can be easily calculated .We could say that the kinetic energy developed  by the electron must be equal to the potential difference between the electron so the equation can written as follows .

1/2 mv2=eVa

and solving for the velocity we get

v=(2eVa/m)(1/2)

when this beam of electron enters into the magnetic field strength B than The beam is deflected into a circular path of radius R by the magnetic force and undergoes a centripetal acceleration. This can be expressed as

so it can be represented as the

eVaB=(m*v2)/r

now from the above given problem we should first calculate the magnetic field intensity

B = ((4/5)^3/2 ) * (µ₀)(N)(I)/r

putting the value of all the given quantities

we get

B = ((4/5)^3/2 ) *(4**10-7)*(130)I/(0.15)

=7.79286*10^{-4 *I}

now , from the equation (2) of the question we get ,

R = mv/eB

or, m/e= R*B/v

or, m/e= 7.798x10^-4*i*R/V

or,m/e=7.798x10^-4*i*R/

or,*m/e= 7.79x10^-4 *I*R /

or, e/m =(2*Va)/(7.79 x10^-4)²

or, e/m=3293336.596*Va/I2R2

hence proved...