First slide

Bohr's atom model

Question

The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number = z) is the same as the shortest wavelength of the Balmer series of hydrogen
atom. The value of z is

Moderate

A

2
B

3
C

4
D

6

Solution

Shortest wavelength of Brackett series corresponds to the transition of electron between $n_{1} = 4 and n_{2} = \infty$ and the shortest wavelength of Balmer series corresponds to the transition of electron between, $n_{1} = 2 a n d n_{2} = \infty$ . So,
$(z^{2}) (\frac{13.6}{16}) = (\frac{13.6}{4})$ $(∵ E = \frac{- 13.6 z^{2}}{n^{2}})$
$∴ z^{2} = 4 or z = 2$

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