JEE PHYSICS MCQ: ATOMS

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The wavelength of radiation emitted is

{\lambda _0}

when an electron jumps from the third to second orbit of hydrogen atom. For the electron jump from fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

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Correct option: B

\left( {\frac{{20}}{{27}}} \right){\lambda _0}

Option B (

\left( {\frac{{20}}{{27}}} \right){\lambda _0}

) is correct because it's the wavelength of radiation emitted when an electron jumps from the fourth to the second orbit, calculated using the Rydberg formula.

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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