JEE PHYSICS MCQ: ATOMS

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If wavelength corresponding to 2nd line of Lyman series is

\lambda

, then the wavelength corresponding to last line of Balmer series will be

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Correct option: B

\frac{{32\lambda }}{9}

Option B (

\frac{{32\lambda }}{9}

) is correct because it is the calculated wavelength for the last line of the Balmer series based on the given wavelength of the second line of the Lyman series.

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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