Question 48: Atoms & Nuclei

Question

The first emission of hydrogen atomic spectrum in Lyman series appears at a wavelength of

Accepted Answer

We know that, $$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
$$\Rightarrow \quad \frac{1}{\lambda}=R\left(\frac{1}{1_1^2}-\frac{1}{2^2}\right)=\frac{3 R}{4} \Rightarrow \lambda=\frac{4}{3 R} \mathrm{~cm}$$

Answer

$$\frac{3 R}{4} \mathrm{~cm}^{-1}$$

Answer

$$\frac{4}{3 R} \mathrm{~cm}$$

Answer

$$\frac{7 R}{144} \mathrm{~cm}^{-1}$$

Answer

$$\frac{400}{9 R} \mathrm{~cm}$$

Atoms & Nuclei

The first emission of hydrogen atomic spectrum in Lyman series appears at a wavelength of

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