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IV. Numerical problems · Q11

Q.UV light of wavelength 1800 Å is incident on a lithium surface whose threshold wavelength is 4965 Å. Determine the maximum energy of the electron emitted.

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Step 1. The incident photon's energy is E=12400λ(A˚)=124001800≈6.89 eVE=\dfrac{12400}{\lambda(\text{\AA})}=\dfrac{12400}{1800}\approx6.89\ \text{eV}.

Step 2. The work function corresponding to the given threshold wavelength is ϕ0=124004965≈2.50 eV\phi_0=\dfrac{12400}{4965}\approx2.50\ \text{eV}. …

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