13.6 Ev



The binding energy of the hydrogen atom in its ground state is 13.6 eV. What is the energy when it is in the n = 5 state? StrategyThe amount of energy required to cause a transition from the ground state to the n= 4 state is equal to the difference in the energy between the two states. The energy for a hydrogen atom in the nth stationary state is given by En=(!13.6 eV)n2. SolutionThe energy of a hydrogen atom in the n= 4 state is 1 2. If potential energy of an electron in ground state is -13.6 eV. The energy of next higher state is: a) -27.2 eV b) -6.8 eV c) -3.4 eV d) none of these please help me. If I understand you question, yes, B is the correct answer. The energy of the levels are negative and go to zero when the electron is at infinity; i.e., when the atom is ionized. The NEXT high level would be C after B. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history.

A

54.4 eV

B

6.8 eV

C

13.6 eV

D

24.5 eV

Solution:

For hydrogen atom Z = 1
$therefore, , , , Ionisation, energy, E_H=frac{2pi^2 me^4}{n_2 h_2}, , , , , , , , , , , ...(i)$
For $He^+ ion, (He^+ = 1s^1)$
so, $(He^+ = H)$ ionisation energy
$, , , , , , , , , , , E_{{He}^+}=frac{2pi^2 me^4 Z^2}{n^2 h^2}, , , , , , , , , , , , ...(ii)$
Eq (i)/Eq (ii), we get
$, , , , , , , , , , , , , , , , , E_{{He}^+}= E_H times Z^2$
$, , , , , , , , , , , , , , , , , , , , = 13.6 times 4$ = 54.4eV

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13.6 Ev

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13.6 Ev

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13.6 Ev To Nm

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