A light whose frequency is equal to 6 × 10 14 H z is incident on a metal whose work function is 2 eV . [ h = 6.63 × 10 - 34 J s , 1 eV = 1.6 × 10 - 19 J ] . The maximum energy of the electrons emitted will be

Question

A light whose frequency is equal to 6 × 10 14 H z is incident on a metal whose work function is 2 eV . [ h = 6.63 × 10 - 34 J s , 1 eV = 1.6 × 10 - 19 J ] . The maximum energy of the electrons emitted will be, 2 . 49 eV, 4 . 49 eV, 0 . 49 eV, 5 . 49 eV

MARKS App · Accepted Answer

K E m a x = h ν - ϕ Where h ν = e n e r g y o f i n c i d e n t p h o t o n , ϕ = w o r k f u n c t i o n KE max = 6.6 × 10 - 34 × 6 × 10 14 - 2 × 1.6 × 10 - 19 = 3 . 96 × 10 - 19 - 3.2 × 10 - 19 = 0.76 × 10 - 19 1.6 × 10 - 19 eV = 0 . 475 eV

Search any question & find its solution

Looking for more such questions to practice?