JEE PHYSICS MCQ: ATOMS

1 Medium

1 of 1

Analyzing +4 | -1

Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?

View answer and explanation

Correct option: A

122 \mathrm{~nm}

Option A (

122 \mathrm{~nm}

) is correct because the Balmer series transitions to n=2, and the subsequent Lyman series transition from n=2 to n=1 corresponds to approximately 122 nm.

More Questions from ATOMS

Practice All Chapter MCQs →

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