The wavelength of the first line in the Balme | vedclass

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Question

(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$.The first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$.Thus,$\lambda = \frac{36}{5R}$.The second line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$:$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{4-1}{16} ight) = R \left( \frac{3}{16} ight)$.Thus,$\lambda_2 = \frac{16}{3R}$.Now,express $\lambda_2$ in terms of $\lambda$:$\lambda_2 = \frac{16}{3R} = \frac{16}{3} 	imes \left( \frac{5 \lambda}{36} ight) = \frac{16 	imes 5 \lambda}{3 	imes 36} = \frac{80 \lambda}{108} = \frac{20 \lambda}{27}$.

Element	Co-factor
$A$. $-1$	$(1)$ $-2$
$B$. $1$	$(2)$ $32$
$C$. $3$	$(3)$ $4$
$D$. $6$	$(4)$ $6$
	$(5)$ $-6$

The wavelength of the first line in the Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line?

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