Question 2: Dual Nature of Matter and Radiation

Question

In an electron gun the potential difference between the filament and plate is $$4000 \mathrm{~V}$$. What will be the velocity of electron emitting from the gain?

Accepted Answer

To determine the velocity of the electron emitted from the gun, we need to apply the concept of energy conservation. Specifically, the kinetic energy (KE) gained by the electron will be equal to the electrostatic potential energy (PE) provided by the potential difference (V) between the filament and the plate.
The electrostatic potential energy can be expressed as:
$$ PE = eV $$
where:
$$ e $$ is the charge of the electron, approximately $$ 1.6 	imes 10^{-19} 	ext{ C} $$
$$ V $$ is the potential difference, which is given as $$ 4000 	ext{ V} $$
The kinetic energy of the electron can be expressed as:
$$ KE = \frac{1}{2} mv^2 $$
where:
$$ m $$ is the mass of the electron, approximately $$ 9.1 	imes 10^{-31} 	ext{ kg} $$
$$ v $$ is the velocity of the electron we need to determine.
Equating the kinetic energy to the potential energy, we have:
$$ eV = \frac{1}{2} mv^2 $$
Rearranging the equation to solve for the velocity $$ v $$ gives:
$$ v = \sqrt{\frac{2eV}{m}} $$
Substituting the given values:
$$ v = \sqrt{\frac{2 	imes 1.6 	imes 10^{-19} 	imes 4000}{9.1 	imes 10^{-31}}} $$
$$ v = \sqrt{\frac{1.28 	imes 10^{-15}}{9.1 	imes 10^{-31}}} $$
$$ v = \sqrt{1.41 	imes 10^{15}} $$
$$ v \approx 3.75 	imes 10^7 	ext{ m/s} $$
Therefore, the correct option is:
Option D: $$ 3.75 	imes 10^7 \mathrm{~m} / \mathrm{s} $$

Answer

$$3 	imes 10^3 \mathrm{~m} / \mathrm{s}$$

Answer

$$3.18 	imes 10^7 \mathrm{~m} / \mathrm{s}$$

Answer

$$3.52 	imes 10^7 \mathrm{~m} / \mathrm{s}$$

Answer

$$3.75 	imes 10^7 \mathrm{~m} / \mathrm{s}$$

Dual Nature of Matter and Radiation

In an electron gun the potential difference between the filament and plate is $4000 \mathrm{~V}$. What will be the velocity of electron emitting from the gain?

Choose the correct answer: