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. \(3.0\times 10^{8}~\text{m/s}\)

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. \(\frac{24m}{25hR}\)
2. \(\frac{25hR}{24m}\)
3. \(\frac{25m}{24hR}\)
4. \(\frac{24hR}{25m}\)

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be \(3.57~\text{V}\). The threshold frequency of the material is:
1. \(4\times10^{15}~\text{Hz}\)
2. \(5\times10^{15}~\text{Hz}\)
3. \(1.6\times10^{15}~\text{Hz}\)
4. \(2.5\times10^{15}~\text{Hz}\)

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