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Zorluk: ZorPhotoelectric Effect and Work Function

Light of frequency 8.0×1014 Hz8.0 \times 10^{14}\text{ Hz} illuminates a photosensitive plate, causing photoelectrons to be emitted with a maximum kinetic energy of 1.2 eV1.2\text{ eV}. If the same plate is subsequently illuminated by light of frequency 1.2×1015 Hz1.2 \times 10^{15}\text{ Hz}, what is the stopping potential, in volts, needed to reduce the photoelectric current to zero? (Take h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s} and 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

Cevap: 2.85 V

Cevap

The stopping potential needed to reduce the photoelectric current to zero is 2.85 V.
Using Einstein's photoelectric equation E=W0+KmaxE = W_0 + K_{\text{max}}, the initial photon energy is E1=hf1=6.6×10−34×8.0×10141.6×10−19=3.3 eVE_1 = h f_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}. Given K1=1.2 eVK_1 = 1.2\text{ eV}, the work function of the metal is W0=3.3 eV−1.2 eV=2.1 eVW_0 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}. For the second frequency f2=1.2×1015 Hzf_2 = 1.2 \times 10^{15}\text{ Hz}, the photon energy is E2=6.6×10−34×1.2×10151.6×10−19=4.95 eVE_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}. The new maximum kinetic energy is K2=4.95 eV−2.1 eV=2.85 eVK_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}. Since eVs=Kmaxe V_s = K_{\text{max}}, the stopping potential required to reduce the current to zero is 2.85 V2.85\text{ V}.

Adım Adım Çözüm

1
Calculate the photon energy E1E_1 of the initial light in electron-volts
E1=6.6×10−34×8.0×10141.6×10−19=3.3 eVE_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}
Photon energy is related to frequency by E=hfE = h f.
2
Determine the work function W0W_0 of the photosensitive plate
W0=E1−K1=3.3 eV−1.2 eV=2.1 eVW_0 = E_1 - K_1 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}
By Einstein's photoelectric equation, Kmax=E−W0K_{\text{max}} = E - W_0.
3
Calculate the photon energy E2E_2 for the second light frequency
E2=6.6×10−34×1.2×10151.6×10−19=4.95 eVE_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}
The energy of the second photon is calculated using f2=1.2×1015 Hzf_2 = 1.2 \times 10^{15}\text{ Hz}.
4
Find the maximum kinetic energy K2K_2 and corresponding stopping potential VsV_s
K2=4.95 eV−2.1 eV=2.85 eVK_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}, giving Vs=2.85 VV_s = 2.85\text{ V}
The stopping potential in volts is numerical equal to the maximum kinetic energy expressed in electron-volts (eVs=Kmaxe V_s = K_{\text{max}}).

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