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Silver has a work function, Φ, of 4.64 eV. If a sample of silver is bombarded with 215 nm photons, what will be the wavelength (in meters) of the photo-ejected electrons? (1 eV = 1.60 * 10^-19 J, the mass of an electron is 9.109 x 10^-31 kg, and Planck's constant is 6.626 x 10^-34 Js.)

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Final Answer:

The wavelength of the photo-ejected electrons is 0.0709 nm.

Step-by-step explanation:

The work function, Φ, of silver is 4.64 eV. Work function is the minimum energy required to remove an electron from the metal. When a sample of silver is bombarded with photons of wavelength 215 nm, the photo-electrons are ejected from the metal. The energy of each photon is given by the equation: E = hc/λ, where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (2.998 x 10^8 m/s) and λ is the wavelength of the photon. Substituting the given values in the equation, we get the energy of each photon as 3.48 x 10^-19 J.

As per the law of conservation of energy, the energy of the photon must be equal to the work function of the silver plus the kinetic energy of the photo-ejected electrons. This is given by the equation: K.E. = 3.48 x 10^-19 J - 4.64 eV. Converting the electron volts to joules, we get the kinetic energy of the photo-ejected electrons as 8.78 x 10^-19 J. Kinetic energy of the electrons is given by the equation: K.E. = ½ mv², where m is the mass of the electron (9.109 x 10^-31 kg) and v is the velocity of the electron. Substituting the given values in the equation, we get the velocity of the electron as 8.14 x 10^7 m/s.

Using the equation: λ = c/v, where c is the speed of light (2.998 x 10^8 m/s) and v is the velocity of the electron, we get the wavelength of the photo-ejected electrons as 0.0709 nm.

Therefore, the wavelength of the photo-ejected electrons is 0.0709 nm.

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