JEE PHYSICS MCQ: ATOMS

1 Medium

1 of 1

Application +4 | -1

The difference between longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (Atomic number

Z

) equal to

\Delta \lambda

. The value of the Rydberg constant for the given atom is:

View answer and explanation

Correct option: C

\frac{31}{5} \frac{1}{\Delta \lambda \cdot Z^2}

Option C (

\frac{31}{5} \frac{1}{\Delta \lambda \cdot Z^2}

) is correct because it correctly relates the Rydberg constant to the difference in wavelengths of the Balmer and Lyman series.

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