In an atom of hydrogen,the atom is exposed to a radiation of wavelenght 970.58Å and atom got excited.Find out the maximum number of waves can be obtained during de excitation.

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Vikas TU · Answer

Hiiii The maximum no of waves excited wil be equal to ??? we know that 1/Lambda = Rh (1/n f 2 – 1/n i 2 ) Rh = 1.0973 * 10 -7 1/(970.58 * 10 -10 ) * 1.0973 * 10 -7 = (1/n f 2 – 1/n i 2 ) There will be varable values of nf and ni

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In an atom of hydrogen,the atom is exposed to a radiation of wavelenght 970.58Å and atom got excited.Find out the maximum number of waves can be obtained during de excitation.

In an atom of hydrogen,the atom is exposed to a radiation of wavelenght 970.58Å and atom got excited.Find out the maximum number of waves can be obtained during de excitation.

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In an atom of hydrogen,the atom is exposed to a radiation of wavelenght 970.58Å and atom got excited.Find out the maximum number of waves can be obtained during de excitation.

In an atom of hydrogen,the atom is exposed to a radiation of wavelenght 970.58Å and atom got excited.Find out the maximum number of waves can be obtained during de excitation.

1 Answers

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