Question

Which requires more energy to move an electron?

from n = 3 to n = 4
from n = 2 to n = 3
from n = 1 to n = 2

Answer 1

C. from n = 1 to n = 2

Step-by-step explanation:

A.

`\Delta E=E_(final)-E_(initial)`

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

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

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

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

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

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

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

`\Delta E=-1312[0.25-1]KJ mol^(-1)`

`\Delta E=984 KJ mol^(-1)`

Answer 2

From n=1 to n=2

Step-by-step 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.