Question

The range of visible light is 4000A° to 7000A°. Will photoelectrons be emitted by a copper surface of work function 4.4 eV, when illuminated by visible light? Give the mathematical prove of your answer?

Answer 1

The range of visible light is $4000\mathrm{A}{}^{\circ}$ to $7000\mathrm{A}{}^{\circ}$. Will photoelectrons be emitted by a copper surface of work function $4.4\mathrm{eV}$, when illuminated by visible light? Give the mathematical prove of your answer?

**Solution:**

$$A _ {f} = 4.4\mathrm{eV} = 4.4 \cdot 1.6 \cdot 10^{-19} = 7.049 \cdot 10^{19}J$$ $$\lambda_1 = 400\mathrm{nm}$$ $$\lambda_2 = 700\mathrm{nm}$$

We can determine if the photoelectrons would be emitted by using a frequency of red edge:

$$\nu_{red} = \frac{A_f}{h},$$

Where $h$ is a Planck's constant and $h = 6.67 \cdot 10^{-34}(J \cdot c)$

$$\nu_{red} = \frac{7.049 \cdot 10^{-19}}{6.67 \cdot 10^{-34}} = 1.0568 \cdot 10^{15}(Hz)$$

And the frequency of visible light is

$$\lambda_1 = \frac{c}{\nu_1}$$ $$\nu_1 = \frac{c}{\lambda_1} = \frac{3 \cdot 10^8}{400 \cdot 10^{-9}} = 0.75 \cdot 10^{15}(Hz)$$ $$\lambda_2 = \frac{c}{\nu_2};$$ $$\nu_2 = \frac{c}{\lambda_2} = \frac{3 \cdot 10^8}{700 \cdot 10^{-9}} = 0.4285 \cdot 10^{15}(Hz)$$

So we can see that

$$\nu_2 < \nu_1 < \nu_{red}$$

**Answer:** That means that the photoelectrons wouldn't be emitted from the surface.