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QuestionA photon with wavelength 3.10 nm3.10 \mathrm{~nm} emits an electron with kinetic energy 96.7eV96.7 \mathrm{eV}. Find the binding energy in J.

Studdy Solution

STEP 1

Assumptions1. The wavelength of the photon is 3.10 nm3.10 \, \mathrm{nm} . The kinetic energy of the emitted electron is 96.7 eV96.7 \, \mathrm{eV}
3. The conversion factor from electron volts to Joules is 1 eV=1.602×10−19 J1 \, \mathrm{eV} =1.602 \times10^{-19} \, \mathrm{J}
4. The Planck's constant is h=6.626×10−34 J⋅sh =6.626 \times10^{-34} \, \mathrm{J \cdot s}
5. The speed of light is c=3.00×108 m/sc =3.00 \times10^{8} \, \mathrm{m/s}

STEP 2

First, we need to calculate the energy of the photon using the formulaphoton=hcλ_{\text{photon}} = \frac{hc}{\lambda}where hh is Planck's constant, cc is the speed of light, and λ\lambda is the wavelength of the photon.

STEP 3

Now, plug in the given values for hh, cc, and λ\lambda to calculate the energy of the photon.
photon=(6.626×10−34 J⋅s)(3.00×108 m/s)3.10×10−9 m_{\text{photon}} = \frac{(6.626 \times10^{-34} \, \mathrm{J \cdot s})(3.00 \times10^{8} \, \mathrm{m/s})}{3.10 \times10^{-9} \, \mathrm{m}}

STEP 4

Calculate the energy of the photon.
photon=6.41×10−19 J_{\text{photon}} =6.41 \times10^{-19} \, \mathrm{J}

STEP 5

Next, we need to convert the kinetic energy of the electron from electron volts to Joules using the conversion factor.
kinetic=96.7 eV×1.602×10−19 J/eV_{\text{kinetic}} =96.7 \, \mathrm{eV} \times1.602 \times10^{-19} \, \mathrm{J/eV}

STEP 6

Calculate the kinetic energy of the electron in Joules.
kinetic=1.55×10−17 J_{\text{kinetic}} =1.55 \times10^{-17} \, \mathrm{J}

STEP 7

Now that we have the energy of the photon and the kinetic energy of the electron, we can find the binding energy of the electron. The binding energy is the difference between the energy of the photon and the kinetic energy of the electron.
binding=photon−kinetic_{\text{binding}} =_{\text{photon}} -_{\text{kinetic}}

STEP 8

Plug in the values for photon_{\text{photon}} and kinetic_{\text{kinetic}} to calculate the binding energy.
binding=6.41×10−19 J−1.55×10−17 J_{\text{binding}} =6.41 \times10^{-19} \, \mathrm{J} -1.55 \times10^{-17} \, \mathrm{J}

STEP 9

Calculate the binding energy of the electron.
binding=−.49×−17 J_{\text{binding}} = -.49 \times^{-17} \, \mathrm{J}The negative sign indicates that energy is required to remove the electron from the atom, which is consistent with the concept of binding energy.
The binding energy of the electron is −.49×−17 J-.49 \times^{-17} \, \mathrm{J}.

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