In a hydrogen spectrum,let lambda be the wave | vedclass

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

Question

(A) For the first line of the Lyman series $(n_1=1, n_2=2)$:$\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) For the first line of the Lyman series $(n_1=1, n_2=2)$:$\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}ight) = R \left(1 - \frac{1}{4}ight) = \frac{3R}{4} \implies \lambda = \frac{4}{3R} \quad \dots(1)$For the $3^{	ext{rd}}$ line of the Paschen series $(n_1=3, n_2=6)$:$\frac{1}{\lambda_3} = R \left(\frac{1}{3^2} - \frac{1}{6^2}ight) = R \left(\frac{1}{9} - \frac{1}{36}ight) = R \left(\frac{4-1}{36}ight) = \frac{3R}{36} = \frac{R}{12} \implies \lambda_3 = \frac{12}{R} \quad \dots(2)$For the $2^{	ext{nd}}$ line of the Balmer series $(n_1=2, n_2=4)$:$\frac{1}{\lambda_2} = R \left(\frac{1}{2^2} - \frac{1}{4^2}ight) = R \left(\frac{1}{4} - \frac{1}{16}ight) = R \left(\frac{4-1}{16}ight) = \frac{3R}{16} \implies \lambda_2 = \frac{16}{3R} \quad \dots(3)$The wavelength difference is given by $\lambda_3 - \lambda_2 = a\lambda$:$a\lambda = \frac{12}{R} - \frac{16}{3R} = \frac{36 - 16}{3R} = \frac{20}{3R}$Substituting $\lambda = \frac{4}{3R}$ into the equation:$a \left(\frac{4}{3R}ight) = \frac{20}{3R} \implies a = \frac{20}{4} = 5$

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

In a hydrogen spectrum,let $\lambda$ be the wavelength of the first transition line of the Lyman series. The wavelength difference between the $3^{\text{rd}}$ transition line of the Paschen series and the $2^{\text{nd}}$ transition line of the Balmer series is $a\lambda$,where $a = ........$

Similar Questions

If the longest wavelength in the ultraviolet region of the hydrogen spectrum is $\lambda_{0}$,then the shortest wavelength in its infrared region is:

$A$ hydrogen atom in its ground state absorbs a photon and the electron transitions to the $5^{th}$ orbit. When the excited atom returns to its ground state,it emits photons of various energies. What is the maximum number of spectral lines that can be obtained in this process?

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$,the emitted radiation has a frequency ($R=$ Rydberg constant,$C=$ velocity of light).

The ratio of the wavelength of the last line of the Paschen series to that of the Balmer series is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module