Exams › JEE Advanced › Physics
Correct answer: 600 nm
The plasma (angular) frequency is wₚ = sqrt(N e² /(e0 m)). With N = 4e27, e = 1.6e-19, e0 = 1e-11, m = 1e-30: N e² = 4e27 * (2.56e-38) = 1.024e-10; divide by (e0 m = 1e-41) gives 1.024e31; sqrt gives wₚ approx 3.2e15 rad/s. Wavelength lambda = 2*pi*c/wₚ = (6.283 * 3e8)/3.2e15 approx 5.9e-7 m approx 600 nm.