The first line in the Lyman series has wavele | vedclass

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

Question

(D) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For

Vedclass · Accepted Answer

$\frac{27}{5}\lambda$

Vedclass · Answer

(D) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$. So,$\lambda = \frac{4}{3R}$.For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$. So,$\lambda_B = \frac{36}{5R}$.Now,taking the ratio: $\frac{\lambda_B}{\lambda} = \frac{36/5R}{4/3R} = \frac{36}{5R} 	imes \frac{3R}{4} = \frac{9 	imes 3}{5} = \frac{27}{5}$.Therefore,$\lambda_B = \frac{27}{5}\lambda$.

List$-I$	List$-II$
$(P)$ image	$(1)$ Reaction with phenyl diazonium salt gives yellow dye.
$(Q)$ image	$(2)$ Reaction with ninhydrin gives purple color and it also reacts with $FeCl _3$ to give violet color.
$(R)$ image	$(3)$ Reaction with glucose will give corresponding hydrazone.
$(S)$ image	$(4)$ Lassiagne extract of the compound treated with dilute HCl followed by addition of aqueous $FeCl _3$ gives blood red color.
	$(5)$ After complete hydrolysis, it will give ninhydrin test and it $\text{DOES NOT}$ give positive phthalein dye test.

The first line in the Lyman series has wavelength $\lambda$. The wavelength of the first line in the Balmer series is

Similar Questions

The incentre of the triangle formed by the lines $x+y=1$,$x=1$,and $y=1$ is

The major product of the following reaction is:

The half-life of $Ra^{226}$ is $1620$ years. Calculate the number of atoms that decay in one second in $1 \ g$ of radium (Avogadro number $= 6.023 \times 10^{23}$).

As per the Bohr model, the minimum energy (in $eV$) required to remove an electron from the ground state of a doubly ionized $Li$ atom $(Z=3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module