Question

Sodium has a work function of 2.36eV. If a beam of monochromatic 442-mm light is incident upon a sample sodium, will electrons be ejected? At what speed?

Answer 1

a) Let's first find the threshold frequency for the sodium at which the electrons still can be ejected from the surface of the metal:

$$\phi = hf_0 ,$$

here, $\phi$ is the work function of sodium, $h$ is the Planck constant and $f_0$ is the threshold frequency for the sodium.

Then, we get:

$$f_0 = \dfrac{\phi}{h} = \dfrac{2.36 eV}{4.135 \cdot 10^{-15} eV \cdot s } = 5.707 \cdot 10^{14} Hz.$$

Let's convert the wavelength of the incident light, $\lambda$, into the frequency $f$ (the wavelenght of the incident light should be 442 nm, otherwise the second part of this question has no meaning):

$$f = \dfrac{c}{\lambda} = \dfrac{3 \cdot 10^{8} \dfrac{m}{s}}{442 \cdot 10^{-9} m} = 6.78 \cdot 10^{14} Hz.$$

Since, $f > f_0$, the electrons will be ejected from the surface of the sodium.

b) We can find at what speed the electrons will be ejected from the surface of the sodium from the Einstein formula for the photoelectric effect:

$$KE = hf - \phi = hf - hf_0 = h(f - f_0),$$ $$\dfrac{1}{2}m_ev^2 = h(f - f_0),$$ $$v = \sqrt{\dfrac{2h(f - f_0)}{m_e}},$$ $$v = \sqrt{\dfrac{2 \cdot 6.626 \cdot 10^{-34} J \cdot s \cdot ( 6.78 \cdot 10^{14} Hz - 5.707 \cdot 10^{14} Hz)}{9.1 \cdot 10^{-31} kg}} = 3.95 \cdot 10^5 \dfrac{m}{s}.$$

Answer:

a) Since, $f > f_0$, the electrons will be ejected from the surface of the sodium.

b) $v = 3.95 \cdot 10^5 \dfrac{m}{s}.$