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The photon energy in units of eV for electromagnetic waves of wavelength 2 cm is:
A) $2.5 \times {10^{ - 19}}$
B) $5.2 \times {10^{ - 16}}$
C) $3.2 \times {10^{ - 16}}$
D) $6.2 \times {10^{ - 5}}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: The electromagnetic waves travel in the light beam. The particles in the electromagnetic waves are called the photon. The photon consists of energy that is the multiple of the wavelength electromagnetic wave. These photons fall on the surface of the objects and excite the electrons in the electrons. The energy of a photon depends on the wavelength of the light.

Complete step by step answer:
Given: The wavelength of the electromagnetic waves is $\lambda = 2\;{\text{cm}} = 2\;{\text{cm}} \times \dfrac{{1\;{\text{m}}}}{{100\;{\text{cm}}}} = 0.02\;{\text{m}}$.
The expression to find the energy of the photon is given as,
$E = \dfrac{{hc}}{\lambda }......\left( 1 \right)$
Here, h is the Planck’s constant and its value is $6.62 \times {10^{ - 34}}\;{\text{kg}} \cdot {{\text{m}}^2}/{\text{s}}$, c is the speed of the light in vacuum and its value is $3 \times {10^8}\;{\text{m}}/{\text{s}}$.
Substitute $\lambda = 0.02\;{\text{m}}$, $h = 6.62 \times {10^{ - 34}}\;{\text{kg}} \cdot {{\text{m}}^2}/{\text{s}}$ and $c = 3 \times {10^8}\;{\text{m}}/{\text{s}}$ in the expression (1) to find the energy of the photon.
$E = \dfrac{{\left( {6.62 \times {{10}^{ - 34}}\;{\text{kg}} \cdot {{\text{m}}^2}/{\text{s}}} \right)\left( {3 \times {{10}^8}\;{\text{m}}/{\text{s}}} \right)}}{{0.02\;{\text{m}}}}$
$E = 993 \times {10^{ - 26}}\;{\text{J}}$
$E = \left( {993 \times {{10}^{ - 26}}\;{\text{J}}} \right)\left( {\dfrac{{1\;{\text{eV}}}}{{1.6 \times {{10}^{ - 19}}\;{\text{J}}}}} \right)$
$E = 6.20 \times {10^{ - 5}}\;{\text{eV}}$

Thus, the photon energy of the electromagnetic waves in units of eV is $6.20 \times {10^{ - 5}}\;{\text{eV}}$and the option (D) is the correct answer.

Additional Information: The energy of the photon depends on the frequency of the incident photon. The energy of the photon produces the photoelectric effect, when incident on the surface of an object. The energy of the photon transfers to the electrons of the object.

Note: Be careful in substituting the values of the wavelength because most of the time frequency of the electromagnetic wave is given and the original formula of the photon energy is also in terms of the frequency of the incident photon.