JEE Main 2025 Physics question
An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10⁻¹¹ m while that for the electron in third excited state is 8.48 × 10⁻¹⁰ m. The ratio of the de Broglie wavelengths of electron in the ground state to that in excited state is
Answer: (1)
Solution
From mvr = nh/2π, λ = h/mv = 2πr/n. Ground state (n = 1): λ₁ = 2πr₁. Third excited state (n = 4): λ₄ = 2πr₄/4. So λ₁/λ₄ = 4r₁/r₄ = (4 × 5.3 ×
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account