Question 78: Atoms & Nuclei

Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $$\mathop A\limits^o $$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

Accepted Answer

$${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} ight]$$
For singly-ionized helium atom, Z = 2. For hydrogen atom Z = 1.
For Balmer series n
1
= 2.
For first spectral line of hydrogen :
$${1 \over {6561}} = {R_H} 	imes {(1)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight] = {{5R} \over {36}}$$
$$ \Rightarrow {R_H} = {{36} \over {5 	imes 6561}}$$
For second spectral line of helium,
$${1 \over \lambda } = {R_H} 	imes {(2)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} ight] = {{3{R_H}} \over 4}$$
$$ = {3 \over 4} 	imes {{36} \over {5 	imes 6561}}$$
$$ \Rightarrow \lambda  = 1215\,\mathop A\limits^o $$

Answer

1215 $$\mathop A\limits^o $$

Answer

1640 $$\mathop A\limits^o $$

Answer

2430 $$\mathop A\limits^o $$

Answer

4687 $$\mathop A\limits^o $$

Atoms & Nuclei

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $\mathop A\limits^o $. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

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