JEE PHYSICS MCQ: ATOMS

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Statement I: Balmer series lies in the visible region of electromagnetic spectrum.
Statement II:

\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)

, where

n=3,4,5, \ldots \ldots

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Correct option: A

Statement I is true; Statement II is true; Statement II is the correct explanation for Statement I

Option A (Statement I is true; Statement II is true; Statement II is the correct explanation for Statement I) is correct because the Balmer series, defined by the given formula, falls within the visible region of the electromagnetic spectrum.

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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