Using conservation laws show that an electron cannot absorb a photon completely.
[ Hint : `hv = mc^(2) - m_(0)c^(2)` and `(hv)/(c) = mv`]

Statement-1:An isolated and free electron cannot absorb a photon completely. Statement-2 : All material particles behave wave like when in motion. Statement-3: Light of intensity l , falls normally upon a plane mirror (perfectly reflecting surface ). The pressure exerted is "2l"/c

Statement-1:An isolated and free electron cannot absorb a photon completely. Statement-2 : All material particles behave wave like when in motion. Statement-3: Light of intensity l , falls normally upon a plane mirror (perfectly reflecting surface ). The pressure exerted is "2l"/c

Calculate the threshold freqency of the metalfrom the photoelectronsare emitted with zero velocity when exposed to radiation of wavelength 6800 Å Hint : KE = (1)/(2) mv^(2) hv= hv_(0) + KE, KE=0 v= (c ) /( lambda)

Calculate the threshold freqency of the metal from the photoelectrons are emitted with zero velocity when exposed to radiation of wavelength 6800 Å Hint : KE = (1)/(2) mv^(2) hv= hv_(0) + KE, KE=0 v= (c ) /( lambda)

Calculate the threshold freqency of the metalfrom the photoelectronsare emitted with zero velocity when exposed to radiation of wavelength 6800 Å Hint : KE = (1)/(2) mv^(2) hv= hv_(0) + KE, KE=0 v= (c ) /( lambda)

(i) In the explanations of photoelectric effect, we assume one photon of frequency v collides with an electron and transfer its energy. This leads to the equation for the maximum energy E_(max) of the emitted electron as E_(max)=hv-phi_(0) Where phi_(0) is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron? (ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

In the explanation of photoelectric effect, we assume one photon of frequency v collidies with an electron and transfers its energy. This leads to the equation for the maximum energy E_max of the emitted electron as E_max=hv-phi_0 Where phi_0 is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron?

Statement-1: In Photoelectric effect , electrons absorbing the photon can not be a free electron. Statement -2: A free electron can't absorb a photon completely

Statement-1: In Photoelectric effect , electrons absorbing the photon can not be a free electron. Statement -2: A free electron can't absorb a photon completely