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m_a_m_a [10]
3 years ago
11

Which requires more energy to move an electron?

Chemistry
2 answers:
Mandarinka [93]3 years ago
4 0

Answer:

From n=1 to n=2

Explanation:

Electrons in n=1 are strongly attracted to the nucleus and therefore will require great force to overcome the electrostatic force of attraction to displace them from the energy level to another.

The electrostatic force reduces as you progress to the outer energy levels.

MrRa [10]3 years ago
4 0

<u>Answer:</u>

<em>C. from n = 1 to n = 2</em>

<u>Explanation:</u>

A.

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2 )}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(4^2 )}-\frac {1}{(3^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(16 )}-\frac {1}{(9)}]KJ mol^{-1}

\Delta E=-1312[0.0625-0.111]KJ mol^{-1}

\Delta E=64 KJ mol^{-1}

B.

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2 )}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(3^2 )}-\frac {1}{(2^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9 )}-\frac {1}{(4)}]KJ mol^{-1}

\Delta E=-1312[0.111-0.25]KJ mol^{-1}

\Delta E=182 KJ mol^{-1}

C.

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2 )}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(2^2 )}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(4)}-\frac {1}{(1)}]KJ mol^{-1}

\Delta E=-1312[0.25-1]KJ mol^{-1}

\Delta E=984 KJ mol^{-1}

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