JEE PHYSICS MCQ: ATOMS

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The ionization energy for second He electron is:

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Correct option: C

54.4 eV

Option C (54.4 eV) is correct because the ionization energy for the second electron of Helium (He+) is calculated using the formula 13.6 * Z², where Z is the atomic number (2).

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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

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{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