Light of wavelength 0.6 μm falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0 . 5 V . With the light of wavelength 0.4 μm , stopping potentials is 1 . 5 V . With this data, the value of h e is n × 10 − 15 V s . Find the value of n .

Step-by-step solution

e V = h c λ − W ⇒ V = ( h e ) c λ − W e V 1 = ( h e ) c λ 1 − W e ……....……(1) V 2 = ( h e ) c λ 2 − W e ……………(2) Solving these two equations, we get, h e = λ 1 λ 2 ( V 1 − V 2 ) c ( λ 2 − λ 1 ) = ( 0.6 × 0.4 × 10 − 12 ) ( 1 . 0 ) ( 3 × 10 8 ) ( 0 . 2 × 10 − 6 ) = 4 × 10 − 15   V   s n = 4