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Calculate the binding energy (in eV) of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm.

User MeTTeO
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Answer:

The binding energy of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm, is 4.086 x 10^11 eV.

Step-by-step explanation:

To calculate the binding energy of electrons in aluminum, we can use the equation for the energy of a photon with a given wavelength.

The energy (E) of a photon is given by:

E = hc / λ

Where:

E = energy of the photon (in joules)

h = Planck's constant (approximately 6.626 x 10^-34 J·s)

c = speed of light (approximately 3.00 x 10^8 m/s)

λ = wavelength of the photon (in meters)

First, we need to convert the given wavelength from nanometers to meters:

λ = 304 nm = 304 x 10^-9 m

Now, we can calculate the energy of the photon:

E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (304 x 10^-9 m)

E ≈ (1.9878 x 10^-25 J) / (304 x 10^-9 m)

Now, let's convert the energy from joules to electronvolts (eV). 1 eV is equal to 1.602 x 10^-19 joules.

E (in eV) ≈ (1.9878 x 10^-25 J) / (304 x 10^-9 m) * (1 eV / 1.602 x 10^-19 J)

E (in eV) ≈ (1.2422 x 10^14 eV) / (304 x 10^-9 m)

E (in eV) ≈ 4.086 x 10^11 eV

So, the binding energy of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm, is 4.086 x 10^11 eV.

User Peter Carrero
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