Question

Q. The energy needed to remove an electron from metallic sodium is 2.28ev. Does sodium show photo electric effect for red light having wavelength λ= 678 nm. What is cut off wavelength for photoelectric emission from sodium?

Answer 1

Answer on Question #58651, Physics / Quantum Mechanics

The energy needed to remove an electron from metallic sodium is 2.28 eV. Does sodium show photo electric effect for red light having wavelength $\lambda = 678$ nm. What is cut off wavelength for photoelectric emission from sodium?

Find: $E_{1} - ?\lambda - ?$

Given:

$$A = 2,28 \times 1,6 \times 10^{-19} \text{ J}$$ $$\lambda_{1} = 678 \times 10^{-9} \text{ m}$$ $$h = 6,626 \times 10^{-34} \text{ J} \times \text{s}$$ $$c = 3 \times 10^{8} \text{ m/s}$$

Solution:

Equation of external photoelectric effect:

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

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

A – electron work function of the metal surface,

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

Conditions the exist of photoelectric effect:

$$\frac{\mathrm{hc}}{\lambda} = A (2)$$

Energy of photon:

$$E_{1} = \frac{\mathrm{hc}}{\lambda_{1}} (3)$$ $$\text{Of (3)} \Rightarrow E_{1} = 2,94 \times 10^{-19} \text{ J}$$ $$1 \text{ eV} - 1,6 \times 10^{-19} \text{ J}$$ $$E_{1} - 2,94 \times 10^{-19} \text{ J}$$ $$E_{1} = 1,84 \text{ eV}$$ $$A = 2,28 \text{ eV}$$

If $E_{1} < A$ then sodium does not show photo electric effect for red light having wavelength $\lambda = 678$ nm.

$$\text{Of (2)} \Rightarrow \lambda = \frac{\mathrm{hc}}{A} (4)$$ $$\text{Of (4)} \Rightarrow \lambda = 545 \times 10^{-9} \text{ m}$$

Answer:

sodium does not show photo electric effect for red light having wavelength $\lambda = 678$ nm;

observed the photoelectric effect from sodium for wavelengths less than $545\mathrm{nm}$ ($\lambda < 545\mathrm{nm}$)

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