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Structure of atom | Class 11 Chemistry (India) | Science | Khan ...

We will learn the dual nature of light and electron, quantum mechanics, quantum numbers. ... Particle nature of electromagnetic radiation: Planck's quantum theory ... To explain the photoelectric effect, 19th-century physicists theorized that the oscillating ... The energy of a photon could be calculated using Planck's equation: .

For more information, see Structure of atom | Class 11 Chemistry (India) | Science | Khan ...

Given that

Metals work function

Na: 2.75 eV,  K : 2.30 eV, M_o : 4.17 eV, N_i : 5.15 eV

Wavelength \lambda = 3300 A^{^o} = 3300 * 10^{-10} m

Distance between the laser and photocell = 1 m   

Step 1: Calculating the photoelectric emission of the radiation

              Energy of proton E=h\upsilon\ =\ \frac{hc}{\lambda}

                   Where, h - Planck's constant([math] 6.63), speed of light c = 3*10^{8} m/s   and \upsilon - frequency.

                     E = \frac{6.63 * 10^{-34} * 3 * 10^8}{3300 * 10^{-10} }

                     E = 6 * 10^{-19} joules

                     E = 3.76 eV                                                                  \because      1 eV = 1.6* 10^{-19} J

              Hence, photoelectric emission of the radiation   E = 3.76 eV

Step 2: Metals under photoelectric emission Identification

               The photon energy can be passed to the electron if the photon energy is at least as large as the work function.

                   Na and K:

                        Work function \phi < E photon energy.

                        So, photo electric emission is possible

                   M_o and   N_i :

                      Work function \phi>E photon energy.

                      So, photo electric emission is not possible

Step 3: Intensity of radiation effect              

              The intensity of the radiation increases since the laser bought closer to the photocell (50 cm). But photon

               energy does not depend on the intensity of the light.

              Thus, the energy of incident radiation remains unchanged with a decrease in distance.