Question

An electron from various excited states of a hydrogen atom emits radiation to come to the ground state. Let ${\lambda _n}$, ${\lambda _g}$ ​ be the de Broglie wavelength of the electron in the ${n^{th}}$ state and the ground state respectively. Let ${\Lambda _n}$​ be the wavelength of the emitted photon in the transition from the ${n^{th}}$ state to the ground state. For large $n$, (A, B are constants).
A) ${\Lambda _n}^2 \approx A + B{\lambda _n}^2$
B) ${\Lambda _n}^2 \approx \lambda$
C) ${\Lambda _n} \approx A + \dfrac{B}{{{\lambda _n}^2}}$
D) ${\Lambda _n} \approx A + B{\lambda _n}$

Answer 1

There is a lowest energy an electron can have and it corresponds to the state called the ground state. When the electron or atom has higher energy than this lowest energy, the atom or electron is said to be in a state called an excited state.

Complete step by step answer:
Let’s define all the data given in the question:
${\lambda _n}$=De Broglie wavelength at ${n^{th}}$ state
${\lambda _g}$= De Broglie wavelength at ground state
${\Lambda _n}$= wavelength of emitted photons ${n^{th}}$ state to ground state.
And it is given that $n$is large.
We know, Rydberg formula, $\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Here, $n_1\& n_2$ are the ground state and ${n^{th}}$ state respectively.
$R =$ Rydberg constant
$\lambda =$ Wavelength
We need to find the relation between${\Lambda _n}$ and the given wavelength at ${n^{th}}$ state and ground state.
Applying the values to the Rydberg formula for each states:
For ground state, $n_1 = 1$
For ${n^{th}}$ state, $n_2 = n$
Applying these values to Rydberg formula:
$\Rightarrow \dfrac{1}{\lambda } = R\left( {1 - \dfrac{1}{{{n^2}}}} \right)$
We know, $\lambda = \dfrac{{nh}}{{2\pi }}$
Where, $h =$ Planck's constant
And all other terms are described earlier.
$\Rightarrow n = \dfrac{{2\lambda \pi }}{h}$
Applying the value of $n$ to the formula, we get,
$\Rightarrow \dfrac{1}{\lambda } = R\left( {1 - \dfrac{{{h^2}}}{{{{(2\lambda \pi )}^2}}}} \right)$…………………………………. (Eqn. P)
No we are going to figure out all the constants from the above equation,
Here, R is Rydberg constant, h is Planck's constant, $2$ and $\pi$ are also constants.
$R$ Will be assumed as $K$
$\dfrac{{{h^2}}}{{{{(2\pi )}^2}}}$ will be assumed as $C$
$\lambda$ will become ${\lambda _n}$ for ${n^{th}}$ state
We are figuring out these constants to reduce the formula to a simplified manner.
Re-writing the equation P, we get,
$\Rightarrow \dfrac{1}{\lambda } = R\left( {1 - \dfrac{{{h^2}}}{{{{(2{\lambda _n}\pi )}^2}}}} \right)$
Making the changes as discussed, the constant gets figured out;
$\Rightarrow \dfrac{1}{\lambda } = K\left( {1 - \dfrac{{{1^{}}}}{{C{\lambda _n}^2}}} \right)$
We have already given that, ${\Lambda _n}$ = wavelength of emitted photons ${n^{th}}$ state to ground state.
That is, $\dfrac{1}{{{\Lambda _n}}} = K\left( {1 - \dfrac{{{1^{}}}}{{C{\lambda _n}^2}}} \right)$
$\Rightarrow {\Lambda _n} = K{\left( {1 - \dfrac{{{1^{}}}}{{C{\lambda _n}^2}}} \right)^{ - 1}}$
It is given that, $n$ is large. As $n$ is large, ${\lambda _n}$ will also be large so the value of $\dfrac{1}{{{\lambda _n}}}$will be very small;
and by applying binomial theorem, we get,
$\Rightarrow {\Lambda _n} = K\left( {1 + \dfrac{1}{{C{\lambda _n}^2}}} \right)$
$\Rightarrow {\Lambda _n} = K + \dfrac{K}{C} \times \dfrac{1}{{{\lambda _n}^2}}$
It is given in the question that, for large $n$, (A, B are constants).
That is $K$will become $A$, $\dfrac{K}{C}$ will become $B$
$\Rightarrow {\Lambda _n} = A + \dfrac{B}{{{\lambda _n}^2}}$
So the final answer is Option C

Note: The Bohr model of the atom is an attempt to explain patterns in the way atoms and electrons absorb, retain, and release energy. This structure of atom resembles the solar system with the atomic nucleus as the centre and electrons moving in a circular path similar to planets around the Sun. The Bohr model represented advancement in the understanding of atomic structure and contributed to the development of quantum mechanics.