Question

If P.E. of an electron is -6.8 eV in Hydrogen then find out
1. K.E.
2. total energy
3. the orbit where electron exist
4. radius of the orbit​

Answer 1

Answers:-

1. Kinetic energy is 3.4 eV.
2. Total energy is -3.4 eV.
3. Electron will exist in 2nd orbit.
4. Radius of the orbit is 2.116 Å

We have:-

Potential energy = -6.8 eV

Solution:-

$\sf{ \large{\bold{1.} \: K.E. = - \frac{P.E.}{2}}} \\ \\ \rightarrow\sf \: K.E. = - (\frac{ - 6.8}{2}) \\ \\ \rightarrow\sf \: K.E. = \frac{ \cancel6.8}{ \cancel2} \\ \\ \rightarrow\red{\boxed{\sf \: K.E. = 3.4eV}} \\ \\ \sf{ \large{\bold{2.} \: T.E. = - K.E}} \\ \\ \rightarrow\green{\sf{\boxed{T.E. = - 3.4eV}}}$

$\sf{ \large{\bold{3.} \: E = - 13.6 \times \frac{ {Z}^{2} }{ {n}^{2} }}} \\ \\ \rightarrow \sf{3.4 = - 13.6 \times \frac{ {1}^{2} }{ {n}^{2} }} \\ \\ \rightarrow \sf{n}^{2} = \frac{ - 13.6}{ - 3.4} \\ \\ \rightarrow \sf{n}^{2} = \frac{ \cancel{- 13.6}}{\cancel{- 3.4}} \\ \\ \rightarrow \sf{n}^{2} = 4 \\ \\ \rightarrow \sf{n = \sqrt{4}} \\ \\ \rightarrow \: \green{\boxed{\sf{n = 2}}} \\ \\ \large \sf{Therefore - } \\ \\ \underline{ \sf{The \: orbit \: where \: electron \: exist \: is \: {2}^{nd}.}} \\ \\ \sf{ \large{\bold{4.} \: r = 0.529 \times \frac{ {n}^{2} }{ {z}^{2}} \: A}} \\ \\ \rightarrow \sf{r = 0.529 \times \frac{ {2}^{2} }{1} A} \\ \\ \rightarrow \sf{r = 0.529 \times 4A} \\ \\ \rightarrow \pink{ \boxed{\sf{r = 2.116A}}}$

Answer 2

Answer:

$\huge\underline\mathfrak{Solution\::}$

Given that,

P.E.of an electron = -6.8 eV

________________________

1.K.E. = -(P.E./2)

=> K.E. = -(-6.8/2)

=> K.E. = 3.4eV

______________________

2. Total energy = - K.E.

=> Total energy = -3.4eV

______________________

$3.\: \:E = - 13.6 \times \frac{z {}^{2} }{n {}^{2} } \\ - 3.4 = - 13.6 \times \frac{1 {}^{2} }{n {}^{2} } \\ n {}^{2} = \frac{ - 13.6}{3.4} \\ n {}^{2} = 4 \\ n = \sqrt{4} \\ n = 2$

So electron exists in 2nd orbital.

________________________

$4. \: \: r = 0.529 \times \frac{n {}^{2} }{z {}^{2} } A\\ \: \: r = 0.529 \times \frac{2 {}^{2} }{1} A \\ \: \: r = 2.116A$

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