# The ratio of the orbital periods of electrons present in the second shell of hydrogen atom and the third shell of (He^+) ion is: A (8 / 27) B (32 / 27) C (27 / 32) D None of these

## Question

The ratio of the orbital periods of electrons present in the second shell of hydrogen atom and the third shell of ( He^{+}) ion is:

A (8 / 27)

B (32 / 27)

C (27 / 32)

D None of these

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## Answer

To find the ratio of the orbital periods of electrons present in the second shell of a hydrogen atom and the third shell of a ( He^{+} ) ion, we use the relationship for the orbital period:

$$ T \propto \frac{n^{3}}{Z^{2}} $$

Where:

- ( n ) is the principal quantum number (shell)
- ( Z ) is the atomic number (number of protons in the nucleus)

For the **second shell of a hydrogen atom**:

- ( n_1 = 2 )
- ( Z_1 = 1 )

For the **third shell of a ( He^{+} ) ion**:

- ( n_2 = 3 )
- ( Z_2 = 2 )

The ratio of the orbital periods ( \frac{T_1}{T_2} ) is given by:

$$ \frac{T_1}{T_2} = \frac{n_1^3}{Z_1^2} \times \frac{Z_2^2}{n_2^3} $$

Substituting the values:

$$ \frac{T_1}{T_2} = \frac{2^3}{1^2} \times \frac{2^2}{3^3} = \frac{8}{1} \times \frac{4}{27} = \frac{8 \times 4}{27} = \frac{32}{27} $$

Therefore, the correct answer is:

[ \boxed{B} ]

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