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Question: In an experiment to determine e/m using Thomson's method, electrons from the cathode accelerate thro...

In an experiment to determine e/m using Thomson's method, electrons from the cathode accelerate through a potential difference of 1.5 kV. The beam coming out of the anode enters crossed electric and magnetic field of strengths 2×104V/m2{\rm{ }}\times{\rm{ }}{10^4}{\rm{V/m}} , and 8.6×104T8.6{\rm{ }}\times {\rm{ }}{10^{ - 4}}{\rm{T}} respectively, The value of e/m. of the electron will be
A. 1.6×1011C/kg{\rm{ }}1.6{\rm{ }}\times {\rm{ }}{10^{11}}{\rm{C/kg}}
B. 1.7×1011C/kg1.7{\rm{ }}\times {\rm{ }}{10^{11}}{\rm{C/kg}}
C. 1.8×1011C/kg1.8{\rm{ }}\times {\rm{ }}{10^{11}}{\rm{C/kg}}
D. 1.9×1011C/kg1.9{\rm{ }}\times {\rm{ }}{10^{11}}{\rm{C/kg}}

Explanation

Solution

This question is based on the J.J Thomson experiment to find the E/m ratio.
You can calculate the e/m ratio by dividing the charge of the electron by its mass.

Complete step by step answer:
The ratio e/m is called the charge per unit mass. J.J Thomson conducted an experiment on cathode ray particles for calculating e/m ratio then he found out that the ratio is independent of nature of electrode and the gas used in the chamber(which he used for the experiment). J.J Thomson experiment is based on the principle of electrons deflection in the electrical and magnetic field . He used this fact to analyse the experiment.
The change on electron e=1.6×1019Ce = 1.6 \times {10^{ - 19}}{\rm{C}}
The electron’s mass m=9.1×1031Kgm = 9.1 \times {10^{ - 31}}{\rm{Kg}}
Now by using these values, we can easily find out the ratio(e/m)\left( {e/m} \right)
em=1.6×1019C9.1×1031Kg=1.8×1011C/Kg\dfrac{e}{m} = \dfrac{{1.6 \times {{10}^{ - 19}}{\rm{C}}}}{{9.1 \times {{10}^{ - 31}}{\rm{Kg}}}} = 1.8 \times {10^{11}}{\rm{C/Kg}}
The deflection of the cathode rays measures the e/m ratio in the presence of the electrical field. When the ray is accelerated then a force is acted due to the electric field on the ray. Since we know the deflection of rays and the time is taken, then we can easily find the velocity of the rays hence acceleration and when the acceleration is known then the e/m ratio is calculated.
Sometimes the values given are irrelevant to the question, but in these types of cases, we should stick to the basics of the concept. By using the basics, we can solve the problem. If we apply the uniform magnetic field in the path of the electron, then the ratio can also be found.

So, the correct answer is “Option C”.

Note:
Here we used the mass of the electron and charge of the electron to calculate the e/m ratio. You may go wrong when there are given too many unwanted values in the question to confuse you, but you only need two values, i.e. mass and charge values to calculate the e/m ratio.