The energy of an electron in the n^{th} orbit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the Lyman series,the transition occurs to the $n_1 = 1$ orbit from $n_2 = 2, 3, 4, \dots$ levels.Using the Rydberg formula: $\frac{1}{\lambda}

Vedclass · Accepted Answer

$910 \ \mathring{A}, 1213 \ \mathring{A}$

Vedclass · Answer

(A) For the Lyman series,the transition occurs to the $n_1 = 1$ orbit from $n_2 = 2, 3, 4, \dots$ levels.Using the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $R \approx 1.097 	imes 10^7 	ext{ m}^{-1}$.For the longest wavelength (minimum energy transition),$n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4}$.$\lambda_{\max} = \frac{4}{3R} \approx \frac{4}{3 	imes 1.097 	imes 10^7} \approx 1216 \ \mathring{A}$ (approx $1213 \ \mathring{A}$).For the shortest wavelength (maximum energy transition),$n_1 = 1$ and $n_2 = \infty$:$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight] = R$.$\lambda_{\min} = \frac{1}{R} \approx \frac{1}{1.097 	imes 10^7} \approx 912 \ \mathring{A}$ (approx $910 \ \mathring{A}$).Thus,the shortest wavelength is $910 \ \mathring{A}$ and the longest is $1213 \ \mathring{A}$.

The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is expressed as $E_n = \frac{-13.6}{n^2} \text{ eV}$. The shortest and longest wavelengths of the Lyman series are:

Similar Questions

Which one of the following equations is correct for the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$?

An eight-digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is: (in $(7!)$)

$A$ rope of mass $5\,kg$ is moving vertically in a vertical position with an upwards force of $100\,N$ acting at the upper end and a downwards force of $70\,N$ acting at the lower end. The tension at the midpoint of the rope is $............\,N$.

The number of nuclei of a radioactive substance at time $t = 0$ is $1000$,and at time $t = 2 \ s$,it is $900$. Then,the number of nuclei at time $t = 4 \ s$ will be:

The ratio of most probable velocity $(u_{mp})$,average velocity $(u_{av})$,and root mean square velocity $(u_{rms})$ is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module