तरंग दैर्ध्य 4000Å के फोटॉन के लिए ऊर्जा ज्ञात कीजिए (eV में) [`h= 6.63 xx 10^(-34)`J.s तथा `c=3xx10^8` m/s]

If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is : Planck's constant h= 6.63 xx 10^(-34) J.s Speed of light c= 3xx10^8 m//s

If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is : Planck's constant h= 6.63 xx 10^(-34) J.s Speed of light c= 3xx10^8 m//s

Ligth of wavelength 4000A^(@) is incident on a metal surface of work function 2.5 eV . Given h=6.62xx10^(-34) Js,c=3xx10^(8) m//s , the maximum KE of photoelectrons emitted and the corresponding stopping potential are respectively

The stopping potential any material is 1.4Volt.What will its threshold wavelength if the value of incident radiation is 5.6 eV. . [h = 6 .63 xx 10^(-34) Js, C= 3 xx 10^8 m//s, eV = 1.6 xx 10^(-19) J] .