Transition in the Hydrogen Spectrum

In order to determine the transition in the Hydrogen spectrum that would have the same wavelength as the Balmer's transition n=4 to n=2 of the Helium spectrum, we need to understand the energy levels and transitions of both Hydrogen and Helium.

Hydrogen Spectrum and Energy Levels
- The Hydrogen spectrum is a series of lines or wavelengths corresponding to the transitions of electrons between energy levels in the Hydrogen atom.
- The energy levels in Hydrogen are labeled with principal quantum numbers (n=1,2,3,...) where n=1 corresponds to the ground state.
- The Balmer series in Hydrogen corresponds to transitions where the electron starts from an energy level with n>2 and ends in the n=2 energy level.

Helium Spectrum and Energy Levels
- The energy levels in Helium are more complex compared to Hydrogen due to the presence of two electrons.
- Each electron in Helium can occupy one of the energy levels, and the energy levels are labeled with principal quantum numbers (n=1,2,3,...) similar to Hydrogen.
- The transition n=4 to n=2 in the Helium spectrum indicates that one of the electrons starts from the n=4 energy level and ends in the n=2 energy level.

Identifying the Transition
To find the transition in the Hydrogen spectrum that has the same wavelength as the Balmer's transition n=4 to n=2 in the Helium spectrum, we need to consider the energy difference between the initial and final energy levels.

1. Determine the energy difference in the Helium spectrum:
- The energy difference between the n=4 and n=2 energy levels in Helium can be calculated using the formula:
ΔE = E_final - E_initial = ( -13.6 eV / 4^2 ) - ( -13.6 eV / 2^2 )
ΔE = ( -13.6 eV / 16 ) - ( -13.6 eV / 4 )
ΔE = -0.85 eV

2. Identify the transition in the Hydrogen spectrum:
- The energy difference in the Hydrogen spectrum should be equal to -0.85 eV to match the Balmer's transition in the Helium spectrum.
- By using the formula ΔE = -13.6 eV ( 1/n_final^2 - 1/n_initial^2 ), we can solve for the appropriate values of n_final and n_initial.
- Substituting ΔE = -0.85 eV and rearranging the equation, we have:
1/n_final^2 - 1/n_initial^2 = -0.85 eV / -13.6 eV
1/n_final^2 - 1/n_initial^2 = 0.0625

By testing various values of n_final and n_initial, we can find the transition in the Hydrogen spectrum that satisfies the equation.

Conclusion
In conclusion, to find the transition in the Hydrogen spectrum that has the same wavelength as the Balmer's transition n=4 to n=2 in the Helium spectrum, we need to calculate the energy difference in the Helium spectrum and then use the energy difference to find the corresponding transition in the Hydrogen spectrum.