Maximum wavelength in balmer series of hydrogen spectrum is

912

A^{0}

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3636

A^{0}

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656

A^{0}

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8201

A^{0}

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Solution

The correct option is B 3636 $A^{0}$
Maximum wave length is obtained when the electrons transition is from n₁= 3 to n₂= 2

Minimum wave Length is obtained when the electrons transition is from

n₁= 2 to n₂= infinite

As you know, to calculate the wavelengths, we have equation, R[1/n₁² - 1/n₂²]

So for maximum wavelength

R[1/2² - 1/3²] which on simplification gives 6568 angstroms

For minimum wavelength

R[1/2² - 1/infinite] which ultimately gives 3636 angstroms

Hence, Max wavelength is 6568 and min wavelength is 3636.

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