JEE Main 2026 Physics question
In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits in given by rᵢ: = 16:4. The wavelength of photon emitted due to this transition is nm. (Given Rydberg constant = 1.0973×10⁷ /m)
Answer: (C)
Solution
Since r∝n², 16=4² and 4=2² give nᵢ=4, =2. 1/λ=R(1/4−1/16)=3R/16, giving λ=16/(3R)≈486 nm.