Question

Electrons are ejected when sodium metal is irradiated with 400 nm light. what is the velocity of the electrons? (φna = 4.4 x 10-19 j)

Answer 1

Answer: $3.5\cdot 10^5 m/s$

Explanation:

In the photoelectric effect, the energy of the incident photon is used partially to extract the photoelectron from the metal (work function) and part is given to the photoelectron as kinetic energy:

$hf=\phi +K$ (1)

where

hf is the energy of the photon, with h being the Planck constant and f the photon frequency

$\phi$ is the work function of the metal

K is the kinetic energy of the electron

In this problem, the photon hasa wavelength of $\lambda=400 nm=4 \cdot 10^{-7} m$, so its frequency is

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8 m/s}{4\cdot 10^{-7} m}=7.5\cdot 10^{14} Hz$

Now we can use eq.(1) to find the Kinetic energy of the photoelectron

$K=hf-\phi = (6.63\cdot 10^{-34} Js)(7.5\cdot 10^{14} Hz)-4.4\cdot 10^{-19} J)=5.7\cdot 10^{-20} J$

And the kinetic energy of the electron is given by:

$K=\frac{1}{2}mv^2$

From which we can find the electron's velocity:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(5.7\cdot 10^{-20}J)}{9.11\cdot 10^{-31} kg}}=3.5\cdot 10^5 m/s$