JEE PHYSICS MCQ: ATOMS

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Maximum wavelength in Balmer series of hydrogen spectrum is (R : Rydberg constant)

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Correct option: D

\frac{{36}}{R}

Option A (

\frac{4}{R}

) is incorrect. The correct answer is

\frac{{36}}{{5R}}

because it is derived from the Balmer series formula where n1 = 2 and n2 = 3 for the maximum wavelength.

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The de-Broglie wavelength of an electron in the first Bohr orbit is The volume occupied by an atom is greater than the volume of the nucleus by a factor of about For a hydrogen atom, when electron jump from

n

=2

{n}=1

, then wave length of radiation emitted found to be

\lambda_0

. Then, for which transition of electron of a 'He' atom, the wavelength of radiation emitted will be equal to '

\lambda_0

'. Energy of characteristic X-ray is a consequence of Taking Rydberg’s constant

{R_H} = 1.097\,\, \times \,\,{10^7}\,\,m

, first and second wavelength of Balmer series in hydrogen spectrum is When hydrogen atom is in its first excited level, its radius is its ground state radius