Exams › JEE Main › Physics

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λₙ, λ_B be the Broglie wavelength of the electron in the nth state and the wavelength of the emitted photon in the transition from the nth state to the ground state. Let λₙ ∝ n, (A, B are constants)

1. Λₙ ≈ A + B/λₙ²
2. Λₙ ≈ A + Bλₙ
3. Λₙ² ≈ A + Bλₙ²
4. Λₙ² ≈ λₙ

Correct answer: Λₙ ≈ A + B/λₙ²

Solution

For the n->1 transition, 1/Lam_n = R(1 - 1/n^2), so Lam_n = 1/[R(1-1/n^2)] ~= (1/R)(1 + 1/n^2). Since the de Broglie wavelength lam_n is proportional to n, 1/n^2 is proportional to 1/lam_n^2, giving Lam_n ~= A + B/lam_n^2.

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 →