What is the significance of Plank's constant in quantum theory? The stopping-potential for photo electric emission from a metallic surface is `9V,e/m = 1.8 x 10^11 c//kg`. Calculate the maximum velocity of photo-electrons.

The maximum kinetic energy of photo electron emitting' from a metallic surface of work function 1.8 eV is 0.5 eV. The corresponding stopping potential is—

For two monochromatic radiations incident on the same metal surface, the stopping potentials are 1.0 V and 2.0 V. The ratio between the maximum velocities, of the emitted photoelectrons is

Einstein's equation for photoelectric effect is E_("max") = hf - W_(0) , where h = Planck's constant = 6.625 xx 10^(-34) J.s, f = frequency of light incident on metal surface, W_(0) = work function of metal and E_("max") = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency f is less than a minimum value f_(0) or if the wavelength lamda is greater than a maximum value lamda_(0) , the value of E_("max") would be negative, which is impossible. Thus for a particular metal surface f_(0) is the threshold frequency and lamda_(0) is the threshold wavelength for photoelectric emssion to take place. Again if the collector plate is ketp at a negative potential with respect to the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becoes zero, is known as the stopping potential, the photoelectric effect stops for a potential lower than this. [velocity of light = 3xx 10^(8) m.s^(-1) , mass of an electron m = 9.1 xx 10^(-31) kg , charge of an electron, e = 1.6 xx 10^(-19)C The threshold wavelength of photoelectric effect for a metal surface is 4600 Å . Work function of the metal (in eV) is

The stopping potential for photoelectrons from a metal surface is V_(1) when monochromatic light of frequency v_(1) is incident on it. The stopping potential becomes V_(2) when monochromatic light of another frequency is incident on the same metal surface. If h be the Planck's constant and e be the charge of an electron, then the frequency of light in the second case is

An electron is subjected to a potential difference of 180 V. Mass and charge of an electron are 9 xx 10^(-31)kg and 1.6 xx 10^(-19)C , respectively. Find the velocity acquired by the electron.

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second from a light source of power 40 W and wavelength 5893 Å