How much work must be done to pull apart the electron and the proton that make up the Hydrogen atom, if the atom is initially in the state with n = 2

13.6 \times 1.6 \times 10^{- 19} J

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3.4 \times 1.6 \times 10^{- 19} J

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1.51 \times 1.6 \times 10^{- 19} J

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Solution

The correct option is B $3.4 \times 1.6 \times 10^{- 19} J$
The electrostatic P.E. is zero when the electron and proton are far apart from each other. Work done in pulling electron and proton far away from each other
$W = E_{f} - E_{i} = 0 - E_{i} = - (- \frac{13.6}{n^{2}} e V) \Rightarrow W = \frac{13.6}{(2)^{2}} \times 1.6 \times 10^{- 19} J = 3.4 \times 1.6 \times 10^{- 19} J$ .

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How much work must be done to pull apart the electron and the proton that make up the Hydrogen atom, if the atom is initially in the state with n = 2

The correct option is B 3.4×1.6×10−19JThe electrostatic P.E. is zero when the electron and proton are far apart from each other. Work done in pulling electron and proton far away from each other W=Ef−Ei=0−Ei=−(−13.6n2eV)⇒W=13.6(2)2×1.6×10−19J=3.4×1.6×10−19J.