Question

The energy of a photon is given as $3.03 \times {10^{ - 19}}j/atom$. The wavelength of the photon is :
(A). $6.56nm$
(B). $65.6nm$
(C). $0.656nm$
(D) $656nm$

Answer 1

Hint:The wavelength of photon can be find using and combining the formula $c = \gamma \lambda $ and $E = \eta \gamma $ photon is packet of energy which is emitted discontinuously in form of small discrete packet of energy.

Complete step by step answer:
Photons are a type of elementary particles. It is the quantum of electromagnetic radiations such as light radial wades etc.
New according to Planck quoting theory, put forward by Max Planck in 1900, the radiant energy is emitted or absorbed not discontinuously in the form of small discrete pockets of energy. Watch packets if energy is called a quantum. In case of light the quantum of energy is called a photon. The energy of each photon is directly proportional to the frequency of radiation.
I.e. $E = \upsilon $ or $E = \eta \upsilon $…(1)
Where $h$ is a proportionality constant called Planck’s constant and its value is $6.63 \times {10^{ - 34}}$ moreover we know that $c = \upsilon \lambda $ or $\upsilon = \dfrac{c}{\lambda }$…(2)
Here $c$ is speed of light in Bascom and its value is $3 \times {10^8}m{s^{ - 1}}$and $\lambda $ is the wavelength of the proton.
Putting equation (2) in (1) we get
$E = \dfrac{{hc}}{\lambda }$ or $\lambda = \dfrac{{hc}}{E}$
Now the energy of the photon is given as $3.03 \times {10^{ - 19}}$j/atom. Putting all values of $h$,$c$,$s$,$E$ in above equation, we get
$\lambda = \dfrac{{6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{3.03 \times {{10}^{ - 19}}J/atom}}$
$ = 6.56 \times {10^{ - 7}}m$ =$656 \times {10^{ - 9}}m$
Or $\lambda = 656nm$
Hence the correct option is D.


Note:
The energy possessed by one mole of quanta (or photon) i.e. The Avogadro number (No) of quanta is called one Einstein of energy.
1 Einstein of Energy =${N_O}$ $h\upsilon = {N_o}\dfrac{{hc}}{\lambda }$