Answer: The threshold energy of the electrons in aluminium is 393.71 kJ/mol.
Explanation: Threshold energy is the minimum amount of energy which is required to emit a photon from a metal surface.
Mathematically,
[tex]E=N_Ah\nu_o[/tex]
Where,
E = Threshold energy
[tex]N_A[/tex] = Avogadro's constant = [tex]6.022\times 10^{23}atoms/mol[/tex]
h = Planck's Constant = [tex]6.624\times 10^{-34}Js[/tex]
[tex]\nu[/tex] = Frequency of the particle =[tex]9.87\times 10^{14}s^{-1}[/tex]
Putting values in above equation, we get:
[tex]E=(6.022\times 10^{23})atoms/mol\times (6.624\times 10^{-34})Js\times (9.87\times 10^{14})s^{-1}[/tex]
[tex]E=393.71\times 10^3J/mol=393.71kJ/mol[/tex] (Conversion Factor: 1kJ = 1000J)
The value of threshold energy of electrons in aluminum is 39.406 kJ/mol.
Given data:
The lowest frequency for the photoelectric effect is, [tex]f=9.87 \times 10^{14} \;\rm Hz[/tex].
The expression for the threshold energy is,
[tex]E = Nhf ..................................(1)[/tex]
Here,
N is the Avogadro's number and its value is [tex]6.022 \times 10^{23}[/tex].
h is the Planck's constant and its value is [tex]6.63 \times 10^{-34} \;\rm J-s[/tex].
Solving as,
[tex]E = 6.022 \times 10^{22} \times 6.63 \times 10^{-34} \times 9.87 \times 10^{14}\\E=39406.82 \;\rm J/mol= 39.406 \;\rm kJ/mol[/tex]
Thus, the value of threshold energy is 39.406 kJ/mol
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