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Pfund series wavelength of H-atom lies in which part of EM waves:

Option: 1

UV region


Option: 2

Visible region


Option: 3

Infrared region


Option: 4

Both 1 and 3


Answers (1)

best_answer

For Pfund Series

\frac{1}{\lambda ^{max}} =R \left ( \frac{1}{5^{2}}-\frac{1}{6 ^{2}} \right )=R\left ( \frac{11}{25*36} \right )

\lambda ^{max} = \frac{900}{11R}

\frac{1}{\lambda ^{min}} = R\left ( \frac{1}{5^{2}}-\frac{1}{\infty ^{2}} \right ) = \frac{R}{25}

\lambda ^{min} = \frac{25}{R}

when\ \frac{1}{R}= 912A^{\circ}

Wavelength lies in infrared regions

Therefore, Option(3) is correct

Posted by

Irshad Anwar

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