Question

Q. The emission of photoelectorns due to incident photons on the sodium surface stopped when a stopping potential of 4.2V is applied. Find wavelength of incident photons.

Answer 1

Answer on Question #58592, Physics / Quantum Mechanics

The emission of photoelectrons due to incident photons on the sodium surface stopped when a stopping potential of 4.2V is applied. Find wavelength of incident photons.

Find: $\lambda - ?$

Given:

$$U = 4,2 \, \mathrm{V}$$ $$A = 2,36 \times 1,6 \times 10^{-19} \, \mathrm{J}$$ $$e = -1,6 \times 10^{-19} \, \mathrm{C}$$ $$h = 6,626 \times 10^{-34} \, \mathrm{J} \times \mathrm{s}$$ $$c = 3 \times 10^{8} \, \mathrm{m/s}$$

Solution:

Equation of external photoelectric effect:

$$\frac{\mathrm{hc}}{\lambda} = A + \frac{\mathrm{mv}_{\mathrm{max}}^2}{2} (1),$$

where $\frac{\mathrm{hc}}{\lambda}$ – energy of photon,

A – electron work function of the metal surface,

$\frac{\mathrm{mv}_{\mathrm{max}}^2}{2}$ – the maximum kinetic energy of the electron

The condition when the electron not reaches the second cathode:

$$\frac{\mathrm{mv}_{\mathrm{max}}^2}{2} = |e| \, \mathrm{U} (2),$$

where $|e| \, \mathrm{U}$ – work of force of electric field

(2) in (1): $\frac{\mathrm{hc}}{\lambda} = A + |e| \, \mathrm{U}$ (3)

Of (3) $\Rightarrow \lambda = \frac{\mathrm{hc}}{A + |e| \, \mathrm{U}}$ (4)

Of (4) $\Rightarrow \lambda = 189 \times 10^{-9} \, \mathrm{m}$

Answer:

189 nm

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