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nuclear fusion | Development, Processes, Equations, & Facts ...

Nuclear fusion, process by which nuclear reactions between light elements form heavier elements. ... The vast energy potential of nuclear fusion was first exploited in ... In cases where the interacting nuclei belong to elements with low atomic ... of mass M. The formula is B = (Zmp + Nmn − M)c², where mp and mn are the ...

For more information, see nuclear fusion | Development, Processes, Equations, & Facts ...

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De Broglie wavelength (video) | Khan Academy

Bohr model energy levels (derivation using physics) ... Absorption and emission ... that a microscope using electron "matter waves" instead of photon light waves ... or in a situation such as the photoelectric effect, then I see it as a small particle.. . ... As large number of electrons absorb energy, energy absorbed per electron ...

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The Physics of Subatomic Particles

dimension 5 ems . by 2 ems., and finally hit a fluorescent screen on which ... this indicates an atomic diameter of about 3.4 x 10''' m for an argon atom, ... quickly lose their energy and fall into the nucleus, which was obviously not what wa s ... Classical physics stated that the emission and absorption of light and other energ -.

For more information, see The Physics of Subatomic Particles

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Problem Solutions

P is the power density (W m−2) and f is the frequency in Hz. ... The absorbed photons represent 0.794 of the total energy. If this were ... 10 cm by 10 cm photocell (with 100% quantum efficiency) illu- ... nor load resistance, Iν = 5 A. The open-circuit voltage is ... The total light power that falls on the solar image is 4410 W. d.

For more information, see Problem Solutions

Given that

Light intensity I = 10^{-5} W m^{-2}

Surface area of sodium photo-cell A = 2 cm^2 = 2 * 10^{-4} m^2

Number of layers n = 5

Work function of the metal \phi = 2 eV

Step 1: Finding the number of conduction electrons  

             Effective atomic area of a sodium atom A_e = 10^{-20} m^2

                   Number of conduction electrons n_e = n * \frac{A}{A_e}     

                              n_e = 5 * \frac{2 * 10^{-4}}{10^{-20}}

                              n_e = 10^{17}

Step 2: The amount of energy consumed per second of the electron

                    Power of the incident light P = I * A

                                                                P = 10^{-5} * 2 * 10^{-4}

                                                               P = 2* 10^{-9} W

                    Relation between power and energy of photon

                                     P = nE

                      Number of photons per second E=\frac{P}{n_e}  

                               E = \frac{2* 10^{-9}}{10^{17}}

                               E = 2* 10^{-26} J/s

Step 3: Time needed for photoelectric emissions

               Time   t = \frac{\phi}{E}

                         t = \frac{2 ev}{2* 10^{-26}} = \frac{2 * 1.6 * 10^{-19}}{2* 10^{-26}}

                         t =1.67* 10^{7} s

                         t = 0.507 years

The time needed for photoelectric emission is almost half a year, which is not realistic. Thus, the description of the wave is in conflict with the experiment.