Exams › JEE Main › Physics

An electron in a hydrogen atom makes a transition from the third excited level to the second excited level, and then from the second excited level to the first excited level. What is the ratio of the emitted wavelengths λ₁: λ₂ in these two transitions?

1. 7/5
2. 27/20
3. 27/5
4. 20/7

Correct answer: 20/7

Solution

The ratio of emitted wavelengths in transitions between energy levels in a hydrogen atom is inversely proportional to the energy differences between those levels. The energy difference for each transition can be calculated using the Rydberg formula, leading to the ratio of wavelengths being 20/7 for the specified transitions.

Related JEE Main Physics questions

• An electron in one Bohr orbit with principal quantum number n = n1 can produce 3 distinct spectral lines on de-excitation. In another orbit with principal quantum number n = n2, it can produce 6 distinct spectral lines. What is the ratio of the orbital speeds of the electron in these two orbits?
• An electron in a hydrogen atom is found in an excited state with energy equal to -3.4 eV. What is its angular momentum?
• A beam of electrons having energy 12.5 eV is directed at hydrogen gas in the gaseous state at room temperature. The emitted spectrum will contain:
• A hydrogen atom and a Li++ ion are each in their second excited state. If their electronic angular momenta are denoted by L_H and L_Li, and their energies by E_H and E_Li respectively, which statement is correct?
• For a proton–electron system, the interaction potential is V = V0 ln(r/r0), where r0 is a constant. If Bohr’s model is assumed to hold, how does the radius rₙ depend on the principal quantum number n?
• In the hydrogen spectrum, the longest wavelength in the ultraviolet region is 122 nm. The shortest wavelength in the infrared region, rounded to the nearest whole number, is

⚔️ Practice JEE Main Physics free + battle 1v1 →