Answer to Question #135670 in Physics for Tapasya

Question

Photoelectric threshold for a metal is 300 nm. Determine the maximum energy of electrons emitted by a radiation of 200 nm

Answer 1

We can find the maximum kinetic energy of electrons from the formula:

$$KE_{max} = h \nu - W = \dfrac{hc}{\lambda} - W,$$

here, $KE_{max}$ is the maximum kinetic energy of electrons, $h = 6.63 \cdot 10^{-34} \ J \cdot s$ is the Planck's constant, $\nu = \dfrac{c}{\lambda}$ is the frequency of the electromagnetic radiation, $c = 3 \cdot 10^8 \ \dfrac{m}{s}$ is the speed of light, $\lambda = 200 \ nm$ is the length of the electromagnetic radiation and $W$ is the work function.

Let's find the work function:

$$W= h \nu_0 = \dfrac{hc}{\lambda_0},$$

here, $\lambda_0 = 300 \ nm$ is the threshold wavelength.

Then, we get:

$$W= \dfrac{6.63 \cdot 10^{-34} \ J \cdot s \cdot 3 \cdot 10^8 \ \dfrac{m}{s}}{300 \cdot 10^{-9} \ m} = 6.63 \cdot 10^{-19} \ J.$$

Finally, we can find the maximum kinetic energy of electrons:

$KE_{max} = \dfrac{6.63 \cdot 10^{-34} \ J \cdot s \cdot 3 \cdot 10^8 \ \dfrac{m}{s}}{200 \cdot 10^{-9} \ m} - 6.63 \cdot 10^{-19} \ J = 3.315 \cdot 10^{-19} \ J.$

Answer:

$KE_{max} = 3.315 \cdot 10^{-19} \ J.$