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Question

Physics Question on Atoms

The shortest wavelength in the Balmer series is (R=1.097×107m1)(R = 1.097 \times 10^7 \,m^{-1})

A

200nm200\, nm

B

256.8nm256.8\, nm

C

300nm300\, nm

D

364.6nm364.6\, nm

Answer

364.6nm364.6\, nm

Explanation

Solution

Wavelength for Balmer series is 1λ=R(1221n2)\frac{1}{\lambda} = R \left(\frac{1}{2^{2}} -\frac{1}{n^{2}}\right) Here, Rydberg's constant R=1.097×107m1R = 1.097 \times 10^7\, m^{-1} 1λ=41.097×107\therefore \frac{1}{\lambda} = \frac{4}{1.097\times10^{7} } =364.6×109m= 364.6 \times 10^{-9} m =364.6nm= 364.6 \,nm