Answer to Question #138603 in Electric Circuits for Sandralyine

Question

Describe and explain one experiment in which light exhibits a wave-like character and one experiment which illustrates the existence of photons. Light of frequency 50 x 10^14Hz liberates electrons with energy 2.31 x 10^-19J from a certain metallic surface. What is the wavelength of ultraviolet light which liberates electrons of energy 8.93 x 10^-19J from the same surface?

Answer 1

Explanations & Calculations

Wave behavior of light:

• Interference is a major property associated with waves hence an experiments which could prove that quality is needed. Young's double slit experiment is the well known one which proves this quality at the experimental level. Light (of pure wavelength) sent through a pair of slits (made close to each other) makes a numerous number of images of those slits on a screen places opposite to the slits. If not for the quality of interference & diffraction, just an image of that pair of slits could be seen on the screen.

Existence of photons:

• Concept of photons are taken in to considerations where any work or energy related behavior has to be explained with respect to the doer (light in this case or more specifically photon behavior of waves). Light is allowed to fall on a metal plate placed inside a vacuumed glass tube which has the facilities to apply a retarding voltage & measure any current generated within the tube. It's observed that as the light (not all but with some of certain frequency) falls on the metal plate some current is generated which means electrons are ejected from that plate due to the light.

• The given sum is associated with the photon behavior of light. When appropriate light falls & ejects electrons from any surface one photon is responsible for a one ejected electron.
• There is an energy called binding energy (BE) for a given material which has to be overcome by the incoming energy if it wish to eject an electron from that material.
• Any incoming light with energy more than BE, accelerates that ejected electron and increases its kinetic energy.
• Incoming energy = BE + kinetic energy
• Therefore to calculate the Binding energy of the given metal

$\qquad\qquad \begin{aligned} \small hf_0 &= \small BE + Energy\\ \small BE &= \small (6.626\times 10^{-34}) \times50\times 10^{14}-2.31\times10^{-19}\\ &= \small \bold{3.082\times 10^{-18}J} \end{aligned}$

• Therefore the photon energy , the incoming UV radiation must possess is,

$\qquad\qquad \begin{aligned} \small E &= \small 3.082\times 10^{-18}J + 8.93\times 10^{-19}J\\ &= \small 3.975\times 10^{-18}J \end{aligned}$

• Hence the frequency is

$\qquad\qquad \begin{aligned} \small hf_{uv}&= \small 3.975\times 10^{-18}J\\ \small f_{uv} &= \small 5.999\times 10^{15}Hz\\ &\approx \small \bold{60\times 10^{14}Hz} \end{aligned}$