Exams › NEET › Physics

For emission, the wave number of the radiation is given as 1/λ = RZ²[1/n₁² − 1/n₂²]. Calculate the wave number for Z = 2, n₁ = 5, and n₂ = 2.

1. 1/λ = RZ²[1/n₁² − 1/n₂²]
2. 1/λ = RZ[1/n₁² − 1/n₂²]
3. 1/λ = R[1/n₁² − 1/n₂²]
4. 1/λ = ZR[1/n₁² − 1/n₂²]

Correct answer: 1/λ = RZ²[1/n₁² − 1/n₂²]

Solution

The formula for the wave number of emitted radiation is correctly given as 1/λ = RZ²[1/n₁² − 1/n₂²]. This is derived from the Rydberg formula for hydrogen-like atoms, where Z is the atomic number, and n₁ and n₂ are the principal quantum numbers.

Related NEET Physics questions

• The ground state energy of H-atom 13.6 eV. The energy needed to ionize H-atom from its second excited state.
• To explain his theory, Bohr used
• The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be
• For the wavelength λ = 975 Å, calculate the number of spectral lines emitted when an electron transitions from n = 4 to n = 1.
• The ratio of temperatures T₁/T₂ is proportional to the cube of the ratio of principal quantum numbers n₁/n₂. If n₁ = 2n₂, find the ratio T₁/T₂.
• According to the Rydberg formula, calculate the wavelength λ₁ for the transition from n = 4 to n = 3.

⚔️ Practice NEET Physics free + battle 1v1 →