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Sagot :
The binding energy of the electron is 1.237 ×10^(-16) kj/ mol.
The given parameter
wavelength of photon,
λ = 0.999nm
= 0.999x10⁻⁹ m
K.E of emitted photon,
K.E = 940 ev
The binding energy of electron is calculated as follows:
From Einstein's mass defect equation=
ΔE = Δmc²
Also, from Einstein's photo electric equation=
E = Ф + K.E
Where;
Ф is binding energy of electron on metal surface.
The energy of on mole of electron, the emitted is calculated as:
[tex]E = hf =h \frac{c}{λ} = (6.626×10^{-34} ) × \frac{{3×10^{8} }}{0.999×10^{-9} } \\\\E= 1.989×10^{-16}[/tex]
The K.E of emitted electron in Joules is
[tex]K.E = 940 × \frac{1}{2} ×1.602 ×\ 10^{-19} J\\\\ = 0.752 × 10^{-16} J[/tex]
The binding energy is electron is calculated as:
[tex]Ф = E - K.E\\= (1.989 - 0.752) ×10^{-16}\\ =1.237 × 10^{-16} J[/tex]
Question: An X-ray photon of wavelength 0.999nm strikes a surface. The emitted electron has a kinetic energy of 940 eV. What isthe binding energy of the electron in kJ/mol? (KE=1/2mv2; 1 electron volt (eV) = 1.602 x10-19J)
Learn more about binding energy: https://brainly.com/question/15718710
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