Question

If the energy of a photon corresponding to a wavelength of \[6000\mathop A\limits^ \circ \] is \[3.32 \times {10^{ - 19}}J\], the photon energy for a wavelength of \[4000\mathop A\limits^ \circ \] will be:
A. 1.4 eV
B. 4.9 eV
C. 3.1 eV
D. 1.6 eV

Answer 1

Hint: The photon is the basic unit of energy of the electromagnetic wave. Each energy packet of the electromagnetic contains energy in the form of an electric field and the magnetic field. The energy of the photon is inversely proportional to the wavelength of the wave.

Formula used:
\[E = \dfrac{{hc}}{\lambda }\]
where E is the energy of the photon with wavelength \[\lambda \] and $h$ is the Plank’s constant and c is the speed of light in vacuum.

Complete step by step solution:
The electromagnetic wave is the transverse wave which comprises the electric field and magnetic field components which are mutually perpendicular to the propagation of the wave. The electromagnetic wave contains energy as the energy packets, called photons. The photon is the basic unit of energy of the electromagnetic wave. Each energy packet of the electromagnetic contains energy in the form of an electric field and a magnetic field. The energy of the photon is inversely proportional to the wavelength of the wave.

As the energy of the photon is inversely proportional to the wavelength of the photon.
\[E \propto \dfrac{1}{\lambda }\]
\[\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}\]
From the given data,
\[{E_1} = 3.32 \times {10^{ - 19}}J\]
\[\Rightarrow {\lambda _1} = 6000\mathop A\limits^ \circ \]
\[\Rightarrow {\lambda _2} = 4000\mathop A\limits^ \circ \]

Putting the value, we get
\[\dfrac{{3.32 \times {{10}^{ - 19}}J}}{{{E_2}}} = \dfrac{{4000\mathop A\limits^ \circ }}{{6000\mathop A\limits^ \circ }}\]
\[\Rightarrow {E_2} = \left( {3.32 \times {{10}^{ - 19}}J} \right)\left( {\dfrac{{6000\mathop A\limits^ \circ }}{{4000\mathop A\limits^ \circ }}} \right)\]
\[\therefore {E_2} = 4.9 \times {10^{ - 19}}J\]
Hence, the energy of the photon with wavelength \[4000\mathop A\limits^ \circ \] is \[4.9 \times {10^{ - 19}}J\].

Therefore, the correct option is B.

Note: We should be careful about the units of the given physical quantities like the wavelength, speed of light and energy. We must use the numerical values of the physical quantities in respective uniform unit systems. As we increase the wavelength of the photon the energy will decrease and when the wavelength is decreased energy will increase.