Question

shortest and longest wavelength for p fund series​

Answer 1

Answer:

For Lyman series, n1=1.

For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n2=∞.

For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., n2=2.

Answer 2

Answer:

shortest wavelength of pfund series is 2279nm

Explanation:

n2=infinite

n1=5

lamda = 25÷R

=2279

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