The wavelength of radiation emitted is {lambd | vedclass

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

Question

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.F

Vedclass · Accepted Answer

$\frac{20}{27}{\lambda _0}$

Vedclass · Answer

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the transition from $n_2 = 3$ to $n_1 = 2$,the wavelength is ${\lambda _0}$:$\frac{1}{\lambda_0} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9 - 4}{36} ight] = \frac{5R}{36}$.For the transition from $n_2 = 4$ to $n_1 = 2$,let the wavelength be $\lambda$:$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{4} - \frac{1}{16} ight] = R \left[ \frac{4 - 1}{16} ight] = \frac{3R}{16}$.Dividing the two equations:$\frac{\lambda}{\lambda_0} = \frac{5R/36}{3R/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$.Therefore,$\lambda = \frac{20}{27}{\lambda _0}$.

The wavelength of radiation emitted is ${\lambda _0}$ when an electron jumps from the third to the second orbit of a hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom,the wavelength of radiation emitted will be

Similar Questions

The energy levels of the hydrogen spectrum are shown in the figure. There are some transitions $A, B, C, D$ and $E$. Transitions $A, B$ and $C$ respectively represent:

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first member of the Balmer and Paschen series in a hydrogen atom,respectively,then the ratio of the respective frequencies,$f_1 / f_2$,is:

When an electron makes a transition from $n=4$ to $n=2$,the emitted line spectrum will be:

An electron jumps from the $4^{\text{th}}$ orbit to the $2^{\text{nd}}$ orbit of a hydrogen atom. Given Rydberg's constant $R_{H}=10^7 \ m^{-1}$,calculate the frequency in $Hz$ of the emitted radiation. (Take $c=3 \times 10^8 \ m/s$)

The shortest wavelength of the Lyman series of a hydrogen atom is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module