The work function of a silver surface is 4.73 eV. What would happen if the silver is hit with a photon with 2.5 eV of energy?
2 answers:
Answer:
The correct option is;
C) Nothing, the photon just bounces off the surface
Explanation:
The given parameters are;
The work function of the silver surface = 4.73 eV
The amount of energy of the photon = 2.5 eV
The kinetic energy, KE of an electron when ejected from a metal surface is given as follows;
KE = h·f - B·E
Where;
hf = The energy of the photon
B·E = The work function of the metal
Whereby the energy of the photon impacting on the surface of the silver is equal to or larger than the work function of the silver surface, then the energy of the photon can be reach the electrons of the silver so as to be enable the electron to be ejected through the metal surface
However, given that the energy of the photon, 2.5 eV, is less than the work function of the silver surface, 4.73 eV, the kinetic energy of an ejected electron, KE, is less 0, therefore, no electron will be ejected form the surface of the silver and the photon just bounces off the surface.
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Answer : The Lewis-dot structure of is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is,
As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in = 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
Hence, the Lewis-dot structure of is shown below.
