Question

How much energy does the electron have initially in the n=4 excited state?what is the change in energy if the electron from part a now drops to the ground state?

Answer 1

what it looks to be that you found in A was the "initial"...b/c the question asks: 
"how much energy does the electron have 'initially' in the n=4 excited state?" 

"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. 

The answer comes to: −1.36×10^−19 J


You use the same equation for the second part as for part a. 
just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. 

Answer 2

Answer:

Energy  the electron have initially $E_4=-0.85 \rm ev$,

The change in energy if the electron from part a now drops to the ground state is $12.75\rm ev$

Explanation:

Given information:

Electron have initially in the n=4 excited state,

We know,

Energy of an electron is given by,

$E_n=-13.6\rm ev\times\frac{z^2}{n^2}$

On substituting n=4 for excited state energy,

$E=-13.6\rm ev\times\frac{1^2}{4^2}=-0.85\rm ev$

Change in energy from excited state to ground state,

For ground state n=1,

$E=-13.6\rm ev\times z^2\times( \frac{1}{(n_2)^2}-\frac{1}{(n_1)^2})$

On substituting $n_1=1$,$n_2=4$,

$\Delta E=E_4-E_1=-13.6\rm ev\times 1^2\times( \frac{1}{(4)^2}-\frac{1}{(1)^2})=12.75ev$

Hence, energy  the electron have initially $E_4=-0.85 \rm ev$,

NOTE:- $1\rm ev=-1.602\times 10^{-19}joule$

Energy  the electron have initially $E_4=-0.85 \rm ev$,

The change in energy if the electron from part a now drops to the ground state is $12.75\rm ev$

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https://brainly.com/question/13818669?referrer=searchResults