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3 years ago

Idk>?????????????????

Physics

1 answer:

Lemur [1.5K]3 years ago

8 0

Answer:

The correct answer is C

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3 years ago

(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the

crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

$E=\frac {-13.6}{n^2}\ eV$

(a) The energy of the electron in n= 6 excited state is:

$E=\frac {-13.6}{6^2}\ eV$

$E=-0.3778\ eV$

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

$E=\frac {-13.6}{1^2}\ eV$

$E=-13.6\ eV$

$Ratio=\frac {E_6}{E_1}$

$Ratio=\frac {-0.3778}{-13.6}$

Ratio = 0.0278

7 0

3 years ago