Question

# The wave number of the spectral line in the emission spectrum of Hydrogen will be equal to $$\dfrac {8}{9}$$ times the Rydberg's constant if the electron jumps from:

A
n=3 to n=1
B
n=10 to n=1
C
n=9 to n=1
D
n=2 to n=1

Solution

## The correct option is B $$n = 3$$ to $$n = 1$$Wave number of spectral line in emission spectrum of hydrogen,$$\bar{v}=R_H\begin{pmatrix}\dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\end{pmatrix}$$ ...(i)Given, $$\bar{v}=\dfrac{8}{9}R_H$$On putting the value of $$\bar{v}$$ in Eq. (i), we get-$$\dfrac{8}{9}R_H=R_H\begin{pmatrix}\dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\end{pmatrix}$$$$\dfrac{8}{9}=\dfrac{1}{(1)^2}-\dfrac{1}{n^2_2}$$$$\dfrac{8}{9}-1=\dfrac{1}{n^2_2}$$$$\dfrac{1}{3}=\dfrac{1}{n_2} \therefore n_2=3$$Hence, electron jumps from $$n_2=3$$ to $$n_1=1$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More