Question

Calculate the energy of the light having wavelength $45nm$.
(Planck’s constant $h = 6.63 \times {10^{ - 34}}Js$, speed of light $c = 3 \times {10^8}m/s$)
A.$6.67 \times {10^{15}}$
B.$6.67 \times {10^{11}}$
C.$4.42 \times {10^{ - 15}}$
D.$4.42 \times {10^{ - 18}}$

Answer 1

Hint:
Photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it. The photoelectric effect would be attributed to the transfer of energy from the continuous light waves to an electron.

Complete step by step answer:
Einstein defined a quantized unit or quantum of EM energy, which is now called a photon, with an energy proportional to the frequency of EM radiation. In equation form, the photon energy is \[E\; = \;hf\], where E is the energy of a photon of frequency f and h is Planck’s constant. This revolutionary idea looks similar to Planck’s quantization of energy states in blackbody oscillators.
Given that
\[\lambda {\text{ }} = {\text{ }}45{\text{ }}nm{\text{ }} = {\text{ }}45{\text{ }}x{\text{ }}{10^{ - 9}}\;m\]
We need to calculate the energy for the given wavelength

We know that
\[E{\text{ }} = {\text{ }}hc/\lambda \]———-(i)
Where
Planck’s constant $h = 6.63 \times {10^{ - 34}}Js$
Speed of light $c = 3 \times {10^8}m/s$
Solution
Substituting the known values in equation (i) we get
\[E = \;6.62607015 \times {10^{ - 34\;}}X\;3{\text{ }} \times {\text{ }}{10^8}{\text{ }}/\;45{\text{ }}x{\text{ }}{10^{ - 9}}\]
\[E = \;4.42{\text{ }} \times {\text{ }}{10^{ - 18}}\]Joules
Hence energy for the given wavelength is \[4.42{\text{ }} \times {\text{ }}{10^{ - 18}}\]Joules
So the correct answer is D)

Note:Light can have a varied range of values of energy/wavelength/frequency covering what's called the spectrum. We frequently split the complete spectrum into smaller regions and speak about the various “kinds” of light. Radio waves are the smallest amount of energetic photons while gamma rays are the most energetic photons. The photoelectric effect would be attributed to the transfer of energy from the continuous light waves to electrons
-A photon may be a particle representing a quantum of light or other radiation. A photon carries energy proportional to the radiation frequency but has zero mass.