Name: Bohr Model of the Hydrogen Atom
Uploaded: 2019-07-19
Duration: 4 min 50 s
Description: In this video, we will discuss the Bohr model of the hydrogen atom and electron transition in the hydrogen atom. Get in this to explore more.

Question

Explain the J.J. Thomson experiment to determine the specific charge of electron.

Accepted Answer

In 1887, J.J. Thomson, measured the specific charge (e/m) of the cathode ray particles. The specific charge is defined as the charge per unit mass of the particle.

Principle : The fact that the cathode rays (electrons) are deflected by electric and magnetic fields is made use of in this method.

Experimental arrangement

A highly evacuated discharge tube used in this experiment is as shown in Fig. Cathode rays are produced by the discharge between the cathode and the anodes $D_{1} and D_{2}$ .. A thin pencil of cathode ray comes out through fine pin holes in the anode discs. The cathode rays then pass between two parallel metal plates $P_{1} and P_{2}$ and strike the flat face of the tube. This face is coated with suitable fluorescent material. A spot of light is produced at $Q$ But when a potential difference $V$ is applied between $P_{1} and P_{2}$ , the beam is deflected to point $Q_{1}$ By the use of a pair of coils, uniform magnetic field is produced perpendicular to the plane of the paper and outwards through out the region between $P_{1} and P_{2}$

Theory:

1. . Determination of $V$ With a given electric intensity between the plates, the magnetic induction $B$ is adjusted until the beam strikes the screen at the original position $Q$ . Then the downward force $E . e$ due to the electric field is balanced by the force $BeV$ due to magnetic induction where $e$ is the charge of the cathode ray particle and $V$ is the velocity of the cathode rays.

$E . e = BeV or V = \frac{E}{V}$ .............................(1)

2 . Determination of $\frac{e}{m}$ m

Now the magnetic induction is switched off. The deflection $Q Q_{1}$ caused by the electric field alone is measured. At the instant when the cathode rays enter the region between the plates initial velocity in the downward direction $u = 0$

Acceleration along the downward direction

$a = \frac{Ee}{m}$ , where $m$ is the mass of each cathode ray particle.

Time for which the electron moves in the electric field

$t = \frac{l}{V}$ where $l$ is the length of either of the plates and $V$ is the velocity of electron.

The deflection produced on the cathode rays along the downward direction in the electric field is given by

$y = \frac{1}{2} \frac{Ee}{m} {(\frac{l}{V})}^{2}$

Substituting for v from equation (1) in equation (2) we get

$y_{1} = \frac{1}{2} \frac{Ee}{m} (\frac{l^{2}}{E^{2}}) B^{2} = \frac{{el}^{2} B^{2}}{mE}$

The displacement of the spot of light on the screen is given by (Fig)

$y = {ky}_{1}$ , where $k$ is a constant determined by the geometry of the discharge tube. Substituting for $y_{1}$ we get,

$y = k [\frac{{el}^{2} B^{2}}{2 mE}] or \frac{e}{m} = \frac{2 yE}{k^{2} l^{2} B^{2}}$

By substituting the known values in the above relation e/m of an electron can be calculated. The value of e/m calculated using this experiment was found to be $1.76 \times 10^{11} {Ckg}^{- 1}$

Explain the J.J. Thomson experiment to determine the specific charge of electron.

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