Examkin

Soru

Zorluk: OrtaPhotoelectric Effect and Work Function

A metal emitter with a work function of 3.30 eV3.30\text{ eV} is illuminated by monochromatic light of frequency 1.20×1015 Hz1.20 \times 10^{15}\text{ Hz}. If the intensity of the light is doubled while keeping its frequency constant, what is the maximum kinetic energy of the emitted photoelectrons? (h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s}, 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. 1.65 eV1.65\text{ eV}Cevap
  2. B
    3.30 eV3.30\text{ eV}
  3. C
    4.95 eV4.95\text{ eV}
  4. D
    6.60 eV6.60\text{ eV}

Cevap

The maximum kinetic energy of the emitted photoelectrons is 1.65 eV1.65\text{ eV}.
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons depends exclusively on the frequency of incident radiation and the work function of the metal (Kmax=hf−W0K_{\text{max}} = hf - W_0). Here, hf=4.95 eVhf = 4.95\text{ eV} and W0=3.30 eVW_0 = 3.30\text{ eV}, yielding Kmax=1.65 eVK_{\text{max}} = 1.65\text{ eV}. Increasing the light intensity increases the rate of photon arrivals and hence the rate of photoelectron emission, but leaves the kinetic energy of individual electrons unchanged.

Adım Adım Çözüm

1
Calculate the energy of the incident photon in Joules using E=hfE = hf
E=(6.6×10−34 J s)×(1.20×1015 Hz)=7.92×10−19 JE = (6.6 \times 10^{-34}\text{ J s}) \times (1.20 \times 10^{15}\text{ Hz}) = 7.92 \times 10^{-19}\text{ J}
The energy delivered by each quantum of light is proportional to its frequency.
2
Convert the photon energy from Joules to electron-volts (eV)
E=7.92×10−19 J1.6×10−19 J/eV=4.95 eVE = \frac{7.92 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} = 4.95\text{ eV}
Converting to electron-volts allows direct comparison with the given work function.
3
Apply Einstein's photoelectric equation: Kmax=E−W0K_{\text{max}} = E - W_0
Kmax=4.95 eV−3.30 eV=1.65 eVK_{\text{max}} = 4.95\text{ eV} - 3.30\text{ eV} = 1.65\text{ eV}
The maximum kinetic energy equals the excess energy of the photon after overcoming the work function.
4
Evaluate the effect of doubling light intensity at constant frequency
The maximum kinetic energy remains 1.65 eV1.65\text{ eV}.
Light intensity determines the number of photons per second (current), but does not alter individual photon energy or the maximum kinetic energy per photoelectron.

Anahtar Kavram

Independence of photoelectron kinetic energy from light intensity
Bu soruyu puanla

Konular

Photoelectric Effect and Work Function

Bu konudan daha fazla soru