Question

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let ${\mathrm{\lambda }}_{\mathrm{n}},{\mathrm{\lambda }}_{\mathrm{g}}$ be the de Broglie wavelength of the electron in the ${\mathrm{n}}^{\mathrm{th}}$state and the ground state respectively. Let ${\mathrm{A}}_{\mathrm{n}}$ be the wavelength of the emitted photon in the transition from the ${\mathrm{n}}^{\mathrm{th}}$ state to the ground state. For large $\mathrm{n}$, ($\mathrm{A},\mathrm{B}$ are constants)

• A. $\wedge {{}^2}_{\mathrm{n}}\approx \hspace{0.17em}\mathrm{A}+{{\mathrm{B\lambda }}^2}_{\mathrm{n}}$
• B. $\wedge {}^2\mathrm{n}\approx \mathrm{\lambda }$
• C. $\wedge {}_{\mathrm{n}}\approx \mathrm{A}+\left(\mathrm{B}/{{\mathrm{\lambda }}^2}_{\mathrm{n}}\right)$
• D. $\wedge {}_{\mathrm{n}}\approx \mathrm{A}+{\mathrm{B\lambda }}_{\mathrm{n}}$

Answer 1

The correct option is C

$\wedge {}_{\mathrm{n}}\approx \mathrm{A}+\left(\mathrm{B}/{{\mathrm{\lambda }}^2}_{\mathrm{n}}\right)$

The above question can be solved as:

Step 1:

We know from Bohr's Postulates:

Angular velocity$$\begin{gathered} =mvr \\ =\frac{nh}{2\mathrm{\pi }} \end{gathered}$$

Step 2:

And now from De Broglie's Theory

${\mathrm{\lambda }}_{\mathrm{de}\mathrm{broglie}}$$$\begin{gathered} =\frac{h}{p} \\ =\frac{2\mathrm{\pi r}}{h} \end{gathered}$$…….$1$

${\mathrm{\lambda }}_{\mathrm{n}}=$$\frac{2{\mathrm{\pi r}}_{\mathrm{n}}}{n}$ ………………….$2$

from $1$ and $2$ we get

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{g}}=\frac{2{\mathrm{\pi r}}_1}{1} \\ =2{\mathrm{\pi r}}_1 \end{gathered}$$

Step 3:

We know from the Rydberg equation,

$\frac{1}{{\Lambda }_n}=R\left[1-\frac{1}{n^2}\right]$

,$$\begin{gathered} \frac{{\lambda }_g}{{\lambda }_n}=\frac{r_n}{n{r}_1} \\ =n \end{gathered}$$

Now,

$$\begin{gathered} {\Lambda }_n=\frac{n^2}{R\left(n^2-1\right)} \\ =\frac{1}{R}\left[\frac{{\lambda }_n^2}{{\lambda }_n^2-{\lambda }_g^2}\right] \end{gathered}$$

$=\frac{1}{R}\left[\frac{1}{1-{\left({\displaystyle \frac{{\lambda }_g}{{\lambda }_n}}\right)}^2}\right]$

$\simeq \frac{1}{R}\left[1+{\left(\frac{{\lambda }_g}{{\lambda }_n}\right)}^2\right]$

${\Lambda }_n\mathit{}\simeq A+\frac{B}{{\lambda }_n^2}$

Therefore, option C is correct.