Single Choice

What would be maximum wavelength for Brackett series of hydrogen-spectrum?

A$$74483\mathring { A } $$
B$$22790\mathring { A } $$
C$$40519\mathring { A } $$
Correct Answer
D$$18753\mathring { A } $$

Solution

Wavelength: $$\dfrac{1}{\lambda} = R \bigg( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \bigg) Z^2$$ where, $$R = 109737.3$$ $$cm^{-1}$$
For hydrogen spectrum, $$Z=1$$
For maximum wavelength of Brackett series, $$n_1 = 4$$ and $$n_2 = 5$$
$$\therefore$$ $$\dfrac{1}{\lambda_m} = 109737.3\times \bigg( \dfrac{1}{4^2} - \dfrac{1}{5^2} \bigg) 1^2$$ $$cm^{-1}$$
$$\implies$$ $$\lambda_m = 0.00040519$$ cm $$ = 40519$$ $$A^o$$

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