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Question:

The wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen is 1.523×10⁶ m⁻¹. Enter 1 if the statement is True or 0 if False.

Solution:

The wave number is obtained by the following expression
ν̄=R(1/n₂² - 1/n₁²)=1.097×10⁷/m ×(1/2² - 1/3²)=1.523×10⁶/m