K.E of photon electrons is `E`. When light of wavelength,`lambda` falls on them.If `K.E` of photon electrons reduces to half i.e.`E/2`, then,wavelength of incident light will be ..

The K.E of the electron is E when the incident wavelength is lamda . To increase the K.E of the electron to 2E, the incident wavelength must be

The kinetic energy of an electron is E when the incident wavelength is lamda To increase ti KE of the electron to 2E, the incident wavelength must be

The kinetic energy of an electron is E when the incident wavelength is lamda To increase ti KE of the electron to 2E, the incident wavelength must be

Light of wavelength 5000Å falls on a sensitive plate with photoelectric work function of 1.9 eV. The K.E. of the photo electron emitted will be

The ratio of wavelength of a photon and that of an electron of same energy E will be

The ratio of wavelength of a photon and that of an electron of same energy E will be

The wavelength lambda_(e ) of an photon of same energy E are related by

K.E. of photo electron is E when incident frequency is lambda_(1) . It is 2E when incident frequency is lambda_(2) . The relation between the wavelength is

Initially a photon of wavelength lambda_(1) falls on the photocathode and emits of maximum energy E_(1) .If the wavelength of the incident photon is changed to lambda_(2) the maximum energy of the electron emitted becomes E_(2) .Then value of he (h=Planck's constant,c= velocity of light) is