Single Choice

In the spectrum of singly ionized helium, the wavelength of a line observed is almost the same as the first line of Balmer series of hydrogen. It is due to transition of electron

AFrom $$n_1= 6 $$ to $$n_2 = 4$$
Correct Answer
BFrom $$n_1= 5$$ to $$n_2 = 3$$
CFrom $$n_1= 4$$ to $$n_2 = 2$$
DFrom $$n_1= 3$$ to $$n_2 = 2$$

Solution

Same wavelength for $$He(Z=2)$$ and $$H(Z=1)$$ atoms
First line of Balmer series
$$(n=3)\rightarrow (n=2)$$
$$\lambda_H = \lambda_He$$
$$ 1^2(\cfrac{1}{2^2}-\cfrac{1}{3^2}) = 2^2 (\cfrac{1}{n_1 ^2}- \cfrac{1}{n_2 ^2})$$
$$ (\cfrac{1}{n_1 ^2}- \cfrac{1}{n_2 ^2}) = \cfrac{5}{36}$$

Positive Integral solutions possible are: $$n_1 =6, n_2 =4$$

SIMILAR QUESTIONS

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