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:

Consider \(3^{\text{rd}}\) orbit of \(He^{+}\) (Helium). Using a non-relativistic approach, the speed of the electron in this orbit will be: (given \(Z=2\) and \(h\) (Planck's constant)\(= 6.6\times10^{-34}~\text{J-s}\))
1. \(2.92\times 10^{6}~\text{m/s}\)
2. \(1.46\times 10^{6}~\text{m/s}\)
3. \(0.73\times 10^{6}~\text{m/s}\)
4. \(3.0\times 10^{8}~\text{m/s}\)

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

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. \(\frac{27}{20}\)

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be:
(\(m\) is the mass of hydrogen atom, \(R\) is Rydberg constant and \(h\) is Plank’s constant)
1. \(\dfrac{24m}{25hR}\)
2. \(\dfrac{25hR}{24m}\)
3. \(\dfrac{25m}{24hR}\)
4. \(\dfrac{24hR}{25m}\)

The energy of a hydrogen atom in the ground state is \(-13.6\) eV. The energy of a \(\mathrm{He}^{+}\) ion in the first excited state will be:
1. \(-13.6\) eV
2. \(-27.2\) eV
3. \(-54.4\) eV
4. \(-6.8\) eV