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Structure of Atom - Discovery of Electron-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The electron was the first subatomic particle discovered, primarily through J.J. Thomson's Cathode Ray Tube (CRT) experiments in 1897.

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Cathode rays are streams of fast-moving electrons originating from the cathode (negative electrode) towards the anode (positive electrode) when a high voltage of approximately 10,000 V10,000\text{ V} is applied at a low pressure of 0.01 mm Hg0.01\text{ mm Hg}.

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Cathode rays travel in straight lines unless acted upon by an external field, produce mechanical effects (can rotate a light paddle wheel), and carry a negative charge.

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The specific charge or charge-to-mass ratio (em\frac{e}{m}) of an electron is constant (1.76×1011 C/kg1.76 \times 10^{11}\text{ C/kg}), regardless of the gas used in the tube or the material of the electrodes.

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The mass of an electron is extremely small, approximately 11840\frac{1}{1840} of the mass of a Hydrogen atom, making it negligible compared to the total mass of the atom.

📐Formulae

e=−1.602×10−19 Ce = -1.602 \times 10^{-19}\text{ C}

me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\text{ kg}

Specific Charge(em)=1.758×1011 C/kg\text{Specific Charge} (\frac{e}{m}) = 1.758 \times 10^{11}\text{ C/kg}

me=11837×mHm_e = \frac{1}{1837} \times m_H

Total charge (Q)=n×e\text{Total charge } (Q) = n \times e

💡Examples

Problem 1:

Calculate the number of electrons that would together weigh 1 gram1\text{ gram}.

Solution:

Mass of one electron me=9.1×10−31 kg=9.1×10−28 gm_e = 9.1 \times 10^{-31}\text{ kg} = 9.1 \times 10^{-28}\text{ g}. Let nn be the number of electrons. n×(9.1×10−28)=1n \times (9.1 \times 10^{-28}) = 1 n=19.1×10−28n = \frac{1}{9.1 \times 10^{-28}} n≈1.098×1027n \approx 1.098 \times 10^{27}

Explanation:

To find the count of electrons in a specific mass, we divide the total mass by the mass of a single electron.

Problem 2:

Calculate the charge of one mole of electrons.

Solution:

Charge of one electron e=1.602×10−19 Ce = 1.602 \times 10^{-19}\text{ C}. Avogadro's number NA=6.022×1023N_A = 6.022 \times 10^{23}. Total Charge=NA×e\text{Total Charge} = N_A \times e Total Charge=(6.022×1023)×(1.602×10−19) C\text{Total Charge} = (6.022 \times 10^{23}) \times (1.602 \times 10^{-19})\text{ C} Total Charge≈96485 C\text{Total Charge} \approx 96485\text{ C}

Explanation:

One mole represents 6.022×10236.022 \times 10^{23} particles. Multiplying this by the elementary charge gives the Faraday constant (FF).