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Question: A beam of electromagnetic radiation of intensity 6.4\(\times\)10\(^{−5}W/cm^2\) is composed of wavel...

A beam of electromagnetic radiation of intensity 6.4×\times105W/cm2^{−5}W/cm^2 is composed of wavelength, λ=310nm. It falls normally on a metal (work function φ=2eV) of surface area of 1cm2^2. If one in 103^3 photons ejects an electron, the total number of electrons ejected in 1s is 10x (hc = 1240eV nm, 1eV = 1.6 ×\times 1019^{−19}J), then x is ______.

Explanation

Solution

We all know that a minimum threshold frequency is required to eject an electron from the metal surface and below this threshold frequency the electron will now eject from the metal surface.

Complete step by step answer:
Electromagnetic radiation is a flow of energy in which electrical and magnetic fields vary simultaneously. Radio waves, microwaves, infrared light, visible light, ultraviolet light, X rays and gamma rays are all electromagnetic radiations. Electromagnetic radiations travel in space and vacuum through oscillating electrical and magnetic fields generated by their particles. The theory of electromagnetic radiation was given by Scottish scientist Sir James Clerk Maxwell in the early 1870s. It was experimentally confirmed by German Physicist Heinrich Hertz. Maxwell suggested that when an electrically charged particle moves under acceleration that time alternating electrical and magnetic fields are produced which helps the particle in propagation.

We all know that energy of the photon is given by,
E=hcλE = \dfrac{{hc}}{\lambda }
Here E is the energy of incident radiation, and λ\lambda is the wavelength of the incident radiation.
We will now substitute hc=1240eV-nm, λ=310nm to find E.
E=2140  eVnm310  nm =4  eV =4  eV×1.6×1019  C1  eV =6.4×1019  C\begin{array}{l} E = \dfrac{{2140\;{\rm{eVnm}}}}{{310\;{\rm{nm}}}}\\\ = 4\;{\rm{eV}}\\\ = 4\;{\rm{eV}} \times \dfrac{{1.6 \times {{10}^{ - 19}}\;{\rm{C}}}}{{1\;{\rm{eV}}}}\\\ = 6.4 \times {10^{ - 19}}\;{\rm{C}} \end{array}.
Since 4eV > φ=2eV then the photoelectric effect will take place.
Express the equation for power in terms of energy, intensity and area.
NE=IANE = IA
Here, N is the no. of photons per second, I is the intensity, A is the surface area and E is the energy of incident radiation.
N×6.4×1019=6.4×105×1 N=1014\begin{array}{l} N \times 6.4 \times {10^{ - 19}} = 6.4 \times {10^{ - 5}} \times 1\\\ N = {10^{14}} \end{array}
Therefore, no of photons falling per second is 101410^{14}.
We will now write the formula for number of photoelectron ejected per second
n=Nmn = \dfrac{N}{m}
Here n is the no of photoelectrons per second, m is the no of photos needed to emit an electron.
m=10310^3 (Given)
n=1014103 n=1011\begin{array}{l} n = \dfrac{{{{10}^{14}}}}{{{{10}^3}}}\\\ n = {10^{11}} \end{array}
We will now compare n with 10x.
10x=1011{10^x} = {10^{11}}
x=11\Rightarrow x = 11

Note: Electromagnetic waves do not require any medium. They can travel in vacuum or space. Electromagnetic waves are of many types and collectively they form electromagnetic spectrum. Different types of electromagnetic waves differ from one another in wavelength or frequency.