Questions
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
a
2
b
3
c
4
d
6
detailed solution
Correct option is A
Shortest wavelength of Brackett series corresponds to the transition of electron between n1=4 and n2=∞ and the shortest wavelength of Balmer series corresponds to the transition of electron between, n1=2 and n2=∞. So,z213.616=13.64∵E=-13.6z2n2∴ z2=4 or z=2Talk to our academic expert!
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