Question

Find out the ratio of the following for photon ${\left( {{v_{\max }}} \right)_{{\text{Lyman}}}}:{\left( {{v_{{\text{max}}}}} \right)_{{\text{Brackett}}}}$
A.$1:16$
B.$16:1$
C.$4:1$
D.$1:4$

Answer 1

Hint: We have to find the relation between the velocity and the orbit of the electron. Since the electron is in motion it will have kinetic energy so we can use the formula of kinetic energy and this energy will be equal to the energy of the photon. On solving them we will get-
$ \Rightarrow v_{{\text{max}}}^2 = \dfrac{{2hcR{Z^2}}}{{mn_1^2}}$ Where R is Rydberg constant, h is Planck’s constant, c is the speed of light, m is the mass of the electron and v is the velocity of electron and ${n_1}$ is the orbit of the electron. So we can establish a relation between the velocity and the orbit of the electron-$ \Rightarrow {v_{\max }} \propto \dfrac{1}{{{n_1}}}$
Now we can find the ratio of velocity for the Lyman and Bracket series.

Complete step by step answer:
We have to find the ratio of maximum velocity of photons for the Lyman series and Brackett series.
We know that in Lyman series, the electron is transitioned from higher energy state to the energy state of electron ${n_1} = 1$ and in Brackett series, the electron is transitioned from higher energy state to the energy state${n_1} = 4$.
So we will have to find the relation between the maximum velocity of the electron and the orbit of the electron.
Since the electron is moving with a particular velocity so it will also have kinetic energy.
We know that Maximum kinetic energy is gives as-
$ \Rightarrow K.{E_{{\text{max}}}} = \dfrac{1}{2}mv_{{\text{max}}}^2$
Where m is the mass of the electron and v is the velocity of the electron.
We know that the energy of a photon is given as-
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$ Where h is Planck’s constant, c is the speed of light and $\lambda $ is the wavelength of the light
Now the kinetic energy of the electron will be equal to the energy of the photon when the electron is ejected from its one orbit to another orbit.
So we can write-
$ \Rightarrow \dfrac{1}{2}mv_{{\text{max}}}^2 = \dfrac{{hc}}{\lambda }$-- (i)
Now we know that wavelength of emitted photon is given as-
$ \Rightarrow \dfrac{1}{\lambda } = \dfrac{{R{Z^2}}}{{n_1^2}}$ Where, R is Rydberg constant.
Then on substituting this value in eq. (i), we get-
$ \Rightarrow \dfrac{1}{2}mv_{{\text{max}}}^2 = \dfrac{{hcR{Z^2}}}{{n_1^2}}$
On solving we get-
$ \Rightarrow v_{{\text{max}}}^2 = \dfrac{{2hcR{Z^2}}}{{mn_1^2}}$
On simplifying, we get-
$ \Rightarrow {v_{\max }} \propto \dfrac{1}{{{n_1}}}$
Then the ratio of ${\left( {{v_{\max }}} \right)_{{\text{Lyman}}}}:{\left( {{v_{{\text{max}}}}} \right)_{{\text{Brackett}}}}$is-
$ \Rightarrow \dfrac{{{{\left( {{v_{\max }}} \right)}_{{\text{Lyman}}}}}}{{{{\left( {{v_{{\text{max}}}}} \right)}_{{\text{Brackett}}}}}} = \dfrac{{{{\left( {{n_1}} \right)}_{{\text{Brackett}}}}}}{{{{\left( {{n_1}} \right)}_{{\text{Lyman}}}}}}$
On putting the given values, we get-
$ \Rightarrow \dfrac{{{{\left( {{v_{\max }}} \right)}_{{\text{Lyman}}}}}}{{{{\left( {{v_{{\text{max}}}}} \right)}_{{\text{Brackett}}}}}} = \dfrac{4}{1}$

Hence the correct answer is option C.

Note:
The name of the series shown by hydrogen in electromagnetic spectrum are-
Lyman series- which has all the wavelengths in the ultraviolet region of the electromagnetic spectrum.
Balmer series- which has all the wavelengths in the visible region.
Paschen series- which has all the wavelength in the Infrared region.
Brackett series- which has all the wavelengths in the infrared region.
Pfund series- which has all the wavelengths in the infrared region.