The maximum energy of photoelectrons emitted from rubidium when it is illuminated by rays with

| July 1, 2021 | education

The maximum energy of photoelectrons emitted from rubidium when it is illuminated by rays with a wavelength of 317 nm is 2.64 * ^ – 19 J. Determine the work function and the red border of the photoelectric effect for rubidium.

λ = 317 nm = 317 * 10 ^ -9 m.

h = 6.6 * 10 ^ -34 J * s.

Ek = 2.64 * 10 ^ -19 J.

C = 3 * 10 ^ 8 m / s.

Av -?

λcr -?

The energy of the incident photons Eph goes to knocking out electrons from the surface of the metal Av and imparting kinetic energy Ek to them.

Eph = Av + ​​Ek – the law of the photoelectric effect.

Av = Eph – Ek.

The energy of photons Eph is expressed by the formula: Eph = h * C / λ, where h is Planck’s constant, C is the speed of light, λ is the wavelength of photons.

Av = h * C / λ – Ek.

Av = 6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s / 317 * 10 ^ -9 m – 2.64 * 10 ^ -19 J = 3.56 * 10 ^ -19 J …

We express the work function of Av to pull out electrons from the metal surface by the formula: Av = h * C / λcr, where λcr is the red border of the photoelectric effect.

λcr = h * C / Av.

λcr = 6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s / 3.56 * 10 ^ -19 J = 556 * 10 ^ -9 m.

Answer: the work function for rubidium is Av = 3.56 * 10 ^ -19 J, λcr = 556 * 10 ^ -9 m.



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