Electrons of mass m with de- Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ_o) of the emitted X-ray is:

1.  ${\mathrm{\lambda }}_0=\frac{2{\mathrm{mc\lambda }}^2}{\mathrm{h}}$

2.  ${\mathrm{\lambda }}_0=\frac{2\mathrm{h}}{\mathrm{mc}}$

3.  ${\mathrm{\lambda }}_0=\frac{2{\mathrm{m}}^2{\mathrm{c}}^2{\mathrm{\lambda }}^2}{{\mathrm{h}}^2}$

4.  ${\mathrm{\lambda }}_0=\mathrm{\lambda }$

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

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is:
1. \(1:1\)
2. \(1:-1\)
3. \(2:-1\)
4. \(1:-2\)

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. \(\frac{24hR}{25m}\)