The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
1. \(1\)
2. \(4\)
3. \(0.5\)
4. \(2\)

Given that the value of the Rydberg constant is \(10^{7}~\text{m}^{-1}\), what will be the wave number of the last line of the Balmer series in the hydrogen spectrum?
1. \(0.5 \times 10^{7}~\text{m}^{-1}\)
2. \(0.25 \times 10^{7} ~\text{m}^{-1}\)
3. \(2.5 \times 10^{7}~\text{m}^{-1}\)
4. \(0.025 \times 10^{4} ~\text{m}^{-1}\)

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:

The hydrogen gas with its atoms in the ground state is excited by monochromatic radiation of \(\lambda = 975~\mathring{\text{A}}\). The number of spectral lines in the resulting spectrum emitted will be:
1. \(3\)
2. \(2\)
3. \(6\)
4. \(10\)

The ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is:

Electron in hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited to the first excited state. The ratio of the wavelengths \(\lambda_1:\lambda_2\) emitted in the two cases is:
1. \(\frac{7}{5}\)
2. \(\frac{20}{7}\)
3. \(\frac{27}{5}\)
4. \(\frac{27}{20}\)

The transition from the state \(n=3\) to \(n=1\) in hydrogen-like atoms results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:
1. \(3\rightarrow 2\)
2. \(4\rightarrow 2\)
3. \(4\rightarrow 3\)
4. \(2\rightarrow 1\)