
Some energy levels of a molecule are shown in fig. The ratio of the wavelengths $r = {\lambda _1}/{\lambda _2}$is given by:

$\left( a \right)$ $r = \dfrac{1}{3}$
$\left( b \right)$ $r = \dfrac{4}{3}$
$\left( c \right)$ $r = \dfrac{2}{3}$
$\left( d \right)$ $r = \dfrac{3}{4}$
Answer
139.8k+ views
Hint: we have been given two wavelengths and here are the two transitions. So assuming the corresponding two these two transitions, there is the emission of photons in each case. So firstly we will find the energy of that particular photon then using the relation which we have given in the formula, we will find the ratio between these two wavelengths.
Formula used
The energy of photons,
$E = \dfrac{{hc}}{\lambda }$
Here,
$E$, will be the energy
$h$, will be the planck's constant
$c$, will be the speed of light
$\lambda $, will be the wavelength
Complete Step By Step Solution: As we know the formula for the energy of the photon and it can be written as
$E = \dfrac{{hc}}{\lambda }$
And from here, $\lambda $can be written as
$ \Rightarrow \lambda = \dfrac{{hc}}{E}$
And hence we can say that
$ \Rightarrow \lambda \propto \dfrac{1}{E}$
Now from the figure,
The energy of a photon of wavelength ${\lambda _2}$will be equal to
$ \Rightarrow \vartriangle {E_2} = - E - \left( { - \dfrac{4}{3}} \right)$
On solving the above equation, we get
$ \Rightarrow \vartriangle {E_2} = \dfrac{E}{3}$
Now we will calculate the Energy of the photon of wavelength ${\lambda _1}$ and it will equal to
$ \Rightarrow \vartriangle {E_1} = - E - \left( { - 2E} \right)$
On solving the above equation, we get
$ \Rightarrow \vartriangle {E_1} = E$
So now we will calculate the ratios between the two energies
Therefore, it can be written as
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\vartriangle {E_2}}}{{\vartriangle {E_1}}}$
Now on substituting the values, we get
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\left( {\dfrac{E}{3}} \right)}}{E}$
So we will solve the final above equation to get the required ratios
$ \Rightarrow r = \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{1}{3}$
Therefore, the option $a$will be the correct choice.
Note: According to Einstein's equation $E = m{c^2}$ anything that possesses energy has mass and anything that has mass has energy which is equal to the mass of the particle times square of the speed of light. Photons have $0$ rest mass but they have Energy which is equal to Planck constant times its frequency. So, if the photon has a fixed amount of energy which will be more than zero with zero rest mass but the photon does not exist which has $0$ moving mass so its energy will be zero.
Formula used
The energy of photons,
$E = \dfrac{{hc}}{\lambda }$
Here,
$E$, will be the energy
$h$, will be the planck's constant
$c$, will be the speed of light
$\lambda $, will be the wavelength
Complete Step By Step Solution: As we know the formula for the energy of the photon and it can be written as
$E = \dfrac{{hc}}{\lambda }$
And from here, $\lambda $can be written as
$ \Rightarrow \lambda = \dfrac{{hc}}{E}$
And hence we can say that
$ \Rightarrow \lambda \propto \dfrac{1}{E}$
Now from the figure,
The energy of a photon of wavelength ${\lambda _2}$will be equal to
$ \Rightarrow \vartriangle {E_2} = - E - \left( { - \dfrac{4}{3}} \right)$
On solving the above equation, we get
$ \Rightarrow \vartriangle {E_2} = \dfrac{E}{3}$
Now we will calculate the Energy of the photon of wavelength ${\lambda _1}$ and it will equal to
$ \Rightarrow \vartriangle {E_1} = - E - \left( { - 2E} \right)$
On solving the above equation, we get
$ \Rightarrow \vartriangle {E_1} = E$
So now we will calculate the ratios between the two energies
Therefore, it can be written as
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\vartriangle {E_2}}}{{\vartriangle {E_1}}}$
Now on substituting the values, we get
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\left( {\dfrac{E}{3}} \right)}}{E}$
So we will solve the final above equation to get the required ratios
$ \Rightarrow r = \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{1}{3}$
Therefore, the option $a$will be the correct choice.
Note: According to Einstein's equation $E = m{c^2}$ anything that possesses energy has mass and anything that has mass has energy which is equal to the mass of the particle times square of the speed of light. Photons have $0$ rest mass but they have Energy which is equal to Planck constant times its frequency. So, if the photon has a fixed amount of energy which will be more than zero with zero rest mass but the photon does not exist which has $0$ moving mass so its energy will be zero.
Recently Updated Pages
Average fee range for JEE coaching in India- Complete Details

Difference Between Rows and Columns: JEE Main 2024

Difference Between Length and Height: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

Algebraic Formula

Difference Between Constants and Variables: JEE Main 2024

Trending doubts
A point charge + 20mu C is at a distance 6cm directly class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

A conducting loop carrying a current is placed in a class 12 physics JEE_Main

Physics Average Value and RMS Value JEE Main 2025

Charging and Discharging of Capacitor

Other Pages
Electromagnetic Waves Chapter - Physics JEE Main

Collision - Important Concepts and Tips for JEE

The diffraction effect can be observed in left A right class 12 physics JEE_Main

A transformer is used to light a 100W and 110V lamp class 12 physics JEE_Main

JEE Advanced 2025 Notes

Determine the time spent by a particle in a magnetic class 12 physics JEE_Main
