Question No. 41 Time Lin The kinetic energy of an electron with de-Broglic wavelength of 0.3 nanometer is 0.168 eV 16,8 eV 1.68 eV 2.5 eV Maria: +4/1 DDDDDDDDD! DDDDDDDDDE DDDDDDDDDD ht Mark for Review & Next C lear Resance l itestsenesindalenkostet center start e d 565pcs da & Nech

Find the energy that should be added to an electron of energy 2eV to reduce its de-Broglie wavelength from 1nm to 0.5nm.

The kinetic energy of an electron is 5 eV . Calculate the de - Broglie wavelength associated with it ( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)

When photon of energy 25eV strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy T_(A)eV and de Brogle wavelength lambda_(A) .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.76 eV is T_(B) = (T_(A) = 1.50) eV .If the de broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) then i. (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV III T_(A) = 2.0 eV IV. T_(B) = 3.5 eV

When photon of energy 25eV strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy T_(A)eV and de Brogle wavelength lambda_(A) .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.76 eV is T_(B) = (T_(A) = 1.50) eV .If the de broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) then i. (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV III T_(A) = 2.0 eV IV. T_(B) = 3.5 eV

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1 , that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2) , absorbs short-wavelength light, the photoelectron is ejected and an H_(2) , ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV . No photoelectrons are detected above 6.0 eV . (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J ) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2) . The energy of the incident light is 21.2 eV Considering an energy cycle, ul("determine") the bond energy E_(D)( eV) of H_(2)^(+) to the first decimal place. If you were unable to determine the values for E_(B) and E_(C) , then use 15.0 eV and 5.0 eV for E_(B) and E_(C) , respectively.

The figure below shows a vaccum tube containing electrodes made of different metals, 1 and 2 whose work functions are phi_(1) and phi_(2) The electrodes are illuminated simultaneously. The maximum kinetic energy of photoelectrons reaching plate 2 is 1 eV and maximum kinetic energy of photoelectrons reaching plate 1 is 3 eV. Assume that photoelectron emitted from either plate do not interact with each other phi_(1)=1.5 eV and phi_(2)=0.7 eV. Find wavelength (in nm) of the electromagnetic wave used