Question

The energy of a photon of light with wavelength \[5000\] \[\mathop A\limits^o \] is approximately \[2.5eV\] ,this way the energy of X-ray photon with wavelength 1 \[\mathop A\limits^o \] would be?
A.\[\dfrac{{2.5}}{{{{\left( {5000} \right)}^2}}}eV\]
B.\[2.5 \times 5000{\text{ }}ev\]
C.\[\dfrac{{25}}{{{{\left( {5000} \right)}^2}}}eV\]
D.\[\dfrac{{2.5}}{{5000}}eV\]

Answer 1

Hint : Photon is an energy packet and to calculate energy we use the Planck-Einstein photoelectric equation. First, we have to write the formula in terms of energy, and then substituting its values, we will get our desired answer.

Complete Step By Step Answer:
Einstein’s photoelectric equation says that the energy of the photon is equal to the Planck’s contestant (h) times the frequency of the oscillations of the oscillator.
So, we can write, \[E = h\nu .........(i)\]
Where \[E\]= Energy of photon
\[h\] = Planck’s constant
\[\nu \] = frequency of oscillations
Also, we know that speed of light, frequency of oscillation of atom and wavelength of electromagnetic wave is interrelated by the formula,
\[C = \lambda \times \nu \]
So, \[\nu = \dfrac{C}{\lambda }......(ii)\]
Where, C= speed of light
\[\lambda \] = wavelength
So by substituting value of frequency (equation \[(ii)\] ) in equation \[(i)\], we will get the formula
\[E = h\dfrac{C}{\lambda }.......(iii)\]
So, now we will put the values for photon
Also, we know \[\begin{gathered}
  h = 6.625 \times {10^{ - 34}}Js \\
  C = 3 \times {10^8} \\
  \lambda = 5000 \times {10^{ - 10}}m \\
\end{gathered} \]
Energy of photon,
 \[\begin{gathered}
  E = h\dfrac{C}{\lambda } \\
  2.5ev = \dfrac{{hC}}{{5000 \times {{10}^{ - 10}}}}{\text{ }} \\
  {\text{hC = 2}}{\text{.5ev}} \times {\text{5000}} \times {10^{ - 10}}........(iv) \\
\end{gathered} \]
Now, we will calculate for the X-ray,
Here \[c{\text{ and h }}\]will be constant, wavelength is given in question for X- rats is \[1\] angstrom.
Here simply we will put the value of equation \[(iv)\] in the in our formula to calculate energy
So here it goes, \[E = \dfrac{{hc}}{{{\lambda _{x - ray}}}}\]
\[\begin{gathered}
  E = \dfrac{{2.5 \times 5 \times {{10}^{ - 10}}}}{{1 \times {{10}^{ - 10}}}} \\
   = 2.5 \times 5000{\text{ }}ev \\
\end{gathered} \]
So, the energy of X- ray is equal to \[2.5 \times 5000{\text{ }}ev\]

Therefore, we can say, the correct answer is option B.

Note :
Photon is the basis of quantum physics, it is a tiny particle full of energy. It is also known as light quantum. This concept was given by Einstein using his experiments of photoelectric effects to prove the particle nature of the light.