The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of H$ = 2.18 imes {10^{ - 18}}J\,\,\,{ m{ato}}{{ m{m}}^{ - 1}}$and $h = 6.625 imes {10^{ - 34}}Js$)

Question

The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of H$ = 2.18 	imes {10^{ - 18}}J\,\,\,{m{ato}}{{m{m}}^{ - 1}}$and $h = 6.625 	imes {10^{ - 34}}Js$)

Accepted Answer

$3.08 	imes {10^{15}}{s^{ - 1}}$

Answer

$2.00 	imes {10^{15}}{s^{ - 1}}$

Answer

$1.54 	imes {10^{15}}{s^{ - 1}}$

Answer

$1.03 	imes {10^{15}}{s^{ - 1}}$

atomic-structure

Question: The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of H$ = 2.18 \times {10^{ - 18}}J\,\,\,{\rm{ato}}{{\rm{m}}^{ - 1}}$and $h = 6.625 \times {10^{ - 34}}Js$)

atomic-structure

Question: The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of H$ = 2.18 \times {10^{ - 18}}J\,\,\,{\rm{ato}}{{\rm{m}}^{ - 1}}$and $h = 6.625 \times {10^{ - 34}}Js$)

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