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Question: A light whose frequency is equal to \[6 \times {10^{14}}Hz\;\]is incident on a metal whose work func...

A light whose frequency is equal to 6×1014Hz  6 \times {10^{14}}Hz\;is incident on a metal whose work function is 2 eV(h=6.63×10−34Js,1eV=1.6×10−19J)2\,eV(h = 6.63 \times {10^{ - 34}}Js,1eV = 1.6 \times {10^{ - 19}}J). The maximum energy of electrons emitted will be:
(A) 2.49 eV2.49\,eV
(B) 4.49 eV4.49\,eV
(C)0.49 eV0.49\,eV
(D) 5.49 eV5.49\,eV

Explanation

Solution

To calculate the maximum energy of the emitted radiation, we have to use the relation between energy and incident frequency. It is also called Planck’s law of quantum theory.

FORMULA USED:
E=hvE = hv
where,
E=E = Energy of the absorbed or emitted radiation
h=h = Planck’s Constant and
v=v = Frequency of the incident radiation.

Complete step by step answer:
Given, Frequency of incident radiation:
v=6×1014Hz  v = 6 \times {10^{14}}Hz\;
Planck’s constant:
h=6.63×10−34Jsh = 6.63 \times {10^{ - 34}}Js
and work function:
W=2eVW = 2eV
According to Planck’s Quantum theory, the atoms or molecules emit or absorb energy discontinuously not continuously in the form of small packets called ‘Quanta’. The energy of these radiation emitted or absorbed is directly proportional to the frequency of the incident radiation.
The expression for energy is given by:
E=hvE = hv
Now, we have to calculate the total energy absorbed so we put the values of h and vh\,and\,v in the above expression. The energy will be;
E=6.63×10−34×  6×1014=3.9756×10−19JE = 6.63 \times {10^{ - 34}} \times \,\,6 \times {10^{14}} = 3.9756 \times {10^{ - 19}}J
E=3.9756×10−191.6×10−19=2.49 eVE = \dfrac{{3.9756 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 2.49\,eV
Now, Total absorbed energy == Threshold energy ++ Kinetic energy of photoelectrons
2.49=2+K.E2.49 = 2 + K.E
K.E=2.49−2=0.49 eVK.E = 2.49 - 2 = 0.49\,eV
Hence, the maximum energy of electrons emitted:
0.49eV0.49eV

So, the correct answer is Option C.

ADDITIONAL INFORMATION:
Maxwell’s theory of electromagnetic radiation failed to explain various phenomena like the emission of radiation from hot bodies, the Photoelectric effect, the line spectra of atoms, etc. When a Black body is heated it emits radiation. The range of wavelengths of these radiations is very wide because these wavelengths change as the temperature changes. This type of radiation is called black body radiation.

Note: Photoelectric effect is the ejection of electrons from the metal surface. The minimum frequency required for the ejection of electrons from the surface of the metal is known as the threshold frequency.