Question

The minimum wavelength of photon is \[5000\mathop A\limits^ \circ \], its energy will be
A. 2.5 eV
B. 50 eV
C. 5.48 eV
D. 7.48 eV

Answer 1

Hint:The energy of the photon is directly proportional to the frequency of the photon and inversely proportional to the wavelength of the photon. The velocity of the electromagnetic wave in a particular medium is constant.

Formula used:
\[c = \nu \lambda \]
Here \[\nu \] is the frequency of the electromagnetic wave and \[\lambda \] is the wavelength of the same electromagnetic wave in vacuum.
\[E = h\nu \]
Here E is the energy of the photon, h is the Plank’s constant and \[\nu \] is the frequency of the electromagnetic wave.

Complete step by step solution:
The minimum wavelength of the photon is given as \[5000\mathop A\limits^ \circ \]. The given wavelength is given in the unit of angstrom. We need to change the unit of the wavelength in S.I. unit.

On changing the given unit to S.I. unit, we get the wavelength of the photon as,
\[\lambda = 5000\mathop A\limits^ \circ \]
\[\Rightarrow \lambda = 5000 \times {10^{ - 10}}m\]
\[\Rightarrow \lambda = 5.0 \times {10^{ - 7}}m\]
The energy of the photon is proportional to the frequency of the photon. We have the wavelength of the photon and to get the frequency of the photon we use the relation between the speed, frequency and wavelength.

Using the formula \[c = \nu \lambda \], the frequency of the photon is,
\[\nu = \dfrac{c}{\lambda }\]
\[\Rightarrow \nu = \dfrac{{3 \times {{10}^8}}}{{5 \times {{10}^{ - 7}}}}Hz\]
\[\Rightarrow \nu = 6 \times {10^{14}}Hz\]

To find the energy of the photon we use the formula of the energy which is proportional to the frequency of the photon. We got the frequency of the photon. Using the formula of the energy of the photon, we get
\[E = h\nu \]
Here, h is the Plank’s constant and its value is \[6.626 \times {10^{ - 34}}Js\]in S.I. unit.
\[E = \left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {6 \times {{10}^{14}}} \right)J\]
\[\Rightarrow E = 3.9756 \times {10^{ - 19}}J\]
One eV is equal to \[1.6 \times {10^{ - 19}}J\], so the energy is,
\[E = \dfrac{{3.9756 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}eV\]
\[\therefore E = 2.48\,eV\]
On rounding-off the energy to one decimal place, we get the energy of the photon as 2.5 eV

Therefore, the correct option is A.

Note: The frequency of the electromagnetic wave is the characteristic feature of the electromagnetic wave because it is constant for particular electromagnetic waves which don't change with change in medium in which the electromagnetic wave is travelling.