Question

When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68\times 10\text{ }Jmo{{l}^{-1}}$. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?

Answer 1

Hint: To solve this question, we should first calculate the energy associated with wavelength. We should focus on finding energy associated with one mole of photons and for that we should multiply the answer with the Avogadro number.

Complete step-by-step answer:
We should know that electromagnetic (EM) radiation is a form of energy that is all around us and takes many forms, such as radio waves, microwaves, X-rays and gamma rays. Sunlight is also a form of EM energy, but visible light is only a small portion of the EM spectrum, which contains a broad range of electromagnetic wavelengths.
The energy (E) associated with 300 nanometre photon is given by:$\begin{aligned} & E=\dfrac{hc}{\lambda } \\\ & \,\,\,\,=\dfrac{(6.626\times {{10}^{-34}})(3.0\times {{10}^{8}}m{{s}^{-1}})}{300\times {{10}^{-9}}} \\\ & \,\,\,\,=6.62\times {{10}^{-9}}J \\\ \end{aligned}$ $\begin{aligned} & E=\dfrac{hc}{\lambda } \\\ & \,\,\,\,=\dfrac{(6.626\times {{10}^{-34}})(3.0\times {{10}^{8}}m{{s}^{-1}})}{300\times {{10}^{-9}}} \\\ & \,\,\,\,=6.62\times {{10}^{-9}}J \\\ \end{aligned}$
Now, we will find energy of one mole of photons.
$\begin{aligned} & \,\,\,\,\,=(6.626\times {{10}^{-19}}J)\times (6.022\times {{10}^{23}}mo{{l}^{-1}}) \\\ & \,\,\,\,\,=3.9\times {{10}^{5}}Jmo{{l}^{-1}} \\\ \end{aligned}$
Now, we will find minimum energy needed to remove a mole of electrons from sodium

& =3.9\times {{10}^{5}}Jmo{{l}^{-1}}-1.68\times {{10}^{5}}Jmo{{l}^{-1}} \\\ & \,\,\,\,\,=(3.99-1.68)\times {{10}^{5}}Jmo{{l}^{-1}} \\\ & \,\,\,\,\,=2.31\times {{10}^{5}}Jmo{{l}^{-1}} \\\ \end{aligned}$$ We will find the minimum energy for one mole of electron $$\begin{aligned} & =\dfrac{2.31\times {{10}^{5}}Jmo{{l}^{-1}}}{6.022\times {{10}^{23}}mo{{l}^{-1}}} \\\ & =3.84\times {{10}^{-19}}J \\\ \end{aligned}$$ Now, by using this we will find the wavelength. $$\begin{aligned} & \lambda =\dfrac{hc}{E} \\\ & \,\,\,\,=\dfrac{(6.626\times {{10}^{-34}})(3.0\times {{10}^{8}}m{{s}^{-1}})}{3.84\times {{10}^{-19}}J} \\\ & \,\,\,\,=517nm \\\ \end{aligned}$$ This wavelength falls in the region of green light. So, this is the maximum wavelength that will cause the emission of a photoelectron. Note: We should here understand that electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles, exerting force on them. EM waves carry energy, momentum and angular momentum away from their source particle and can impart those quantities to matter with which they interact. One important thing that we should know about is the effect of electromagnetic radiation upon chemical compounds and biological organisms that depend both upon the radiation's power and its frequency.