Question

When a hydrogen atom is in its second excited state, find the ratio of maximum and minimum wavelength?

Answer 1

Answer:

As total energy of electron in nth orbit of H2 atom is given by :- E= - 13 . 6 n 2 e V . Here, negative sign shows that the electron is bound to the nucleus and is not free to leave it. Therefore,ratio of maximum wavelength/ minimum wavelength = ratio of maximum energy / minimum energy.

Explanation:

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Answer 2

Answer:

The ratio of minimum to maximum wavelength in the Balmer series is

5: 9.

Explanation:

• The Balmer series is a collection of six named series that describe the total hydrogen atoms' spectral line emissions.
• When an electron in a hydrogen atom transits from an n=3 or larger orbital to an n=2 orbital, the lines are emitted.

According to the equation of the Balmer series

1/λ=R(1/n₁²−1/n₂²)

=λmin/λmax=(1/2²−1/3²)/(1/2²−1/∞²)

=5/9

Hence, The ratio of minimum to maximum wavelength in the second excited state is 5: 9.

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