A free electron of 2 . 6 eV energy collides with a H + ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. h = 6 . 6 × 10 - 34 J s

Question

A free electron of 2 . 6 eV energy collides with a H + ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. h = 6 . 6 × 10 - 34 J s, 9 . 0 × 10 27 MHz, 1 . 45 × 10 9 MHz, 0 . 19 × 10 15 MHz, 1 . 45 × 10 16 MHz

MARKS App · Accepted Answer

By Energy conservation K . E . e = T . E . H + + E P h o t o n 2 . 6 = - 13 . 6 4 + h f h f = 6 eV f = 6 × 1 . 6 × 10 - 19 6 . 626 × 10 - 34 f = 1 . 45 × 10 15 Hz = 1 . 45 × 10 9 MHz

Search any question & find its solution

Looking for more such questions to practice?