The energy required to remove an electron from a gaseous atom is called

Physics

1 answer:

IRISSAK [1]4 years ago

6 0

Answer: The energy required to remove an electron from a gaseous atom is called ionization energy.

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.

Smaller is the size of an atom more will be the force of attraction between its protons and electrons. Hence, more amount of energy is required to remove an electron.

Thus, we can conclude that the energy required to remove an electron from a gaseous atom is called ionization energy.

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Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$

Coulombs's Law= $\frac{Kq_{1}q_{2} }{r^{2}}$

6 0

4 years ago

Read 2 more answers

Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i

LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

$v^{2} = u^{2} + 2as$