Question

The approximate wavelength of a photon of energy 2.48 eV is:
A. \[500\mathop A\limits^ \circ \]
B. \[5000\mathop A\limits^ \circ \]
C. \[2000\mathop A\limits^ \circ \]
D. \[1000\mathop A\limits^ \circ \]

Answer 1

Hint: The photon is the qualitative unit of energy of the light wave. It is proportional to the frequency of the light wave. The frequency of the light wave is the most characteristic property because it doesn’t change by changing the medium in which the light is travelling.

Formula used:
\[E = h\nu \]
where h is the Plank’s constant and E is the energy of the photon with a frequency equal to \[\nu \].
\[c = \nu \lambda \]
where c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.

Complete step by step solution:
The energy of the photon is given as 2.48 eV. We need to change the unit of energy from eV to joules. One electron-Volt is the energy used to accelerate an electron in a region of electric potential 1 volt. So,
\[1eV = 1.6 \times {10^{ - 19}}J\]
Hence, the energy of the photon is,
\[E = 2.4 \times \left( {1.6 \times {{10}^{ - 19}}} \right)J\]
\[\Rightarrow E = 3.84 \times {10^{ - 19}}J \\ \]
Using the expression for the energy, we find the frequency of the photon is,
\[\nu = \dfrac{E}{h}\]
\[\Rightarrow \nu = \dfrac{{3.84 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}}Hz\]
\[\Rightarrow \nu = 5.80 \times {10^{14}}Hz \\ \]
We got the frequency of the photon and now we need to use the relation of the characteristic physical quantities for the electromagnetic wave to find the wavelength of the photon,
\[c = \nu \lambda \]
\[\Rightarrow \lambda = \dfrac{c}{\nu } \\ \]
On putting the value of the speed of light and the frequency of the photon, we get
\[\lambda = \dfrac{{3 \times {{10}^8}}}{{5.80 \times {{10}^{14}}}}m\]
\[\Rightarrow \lambda = 5.18 \times {10^{ - 7}}m\]
\[\Rightarrow \lambda = \left( {5.18 \times {{10}^3}} \right) \times {10^{ - 10}}m\]
As we know that \[{10^{ - 10}}m = 1\mathop A\limits^ \circ \\ \]
So, the wavelength of the given photon will be,
\[\lambda = \left( {5.18 \times {{10}^3}} \right)\mathop A\limits^ \circ \]
\[\Rightarrow \lambda = 5180\mathop A\limits^ \circ \]
\[\therefore \lambda \approx 5000\mathop A\limits^ \circ \]
Hence, the approximate value of the wavelength of the given photon is 5000 angstrom.

Therefore, the correct option is B.

Note: We must be careful about the units of the physical quantity while solving numerical problems. We need to convert all the given data in standard unit form. Moreover, the photon's rest mass is zero. The photon is a quantization of electromagnetic wave energy. The characteristic of motion is a particle's momentum.