The shortest wavelength of the Brackett series of 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

Question

The shortest wavelength of the Brackett series of 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) 3 (B) 4 (C) 5 (D) 2

Tardigrade Admin · Accepted Answer

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

and

n_{2} = \infty

So,

Z^{2} (\frac{13.6}{4 ^{2}}) = \frac{13.6}{2 ^{2}}

or

Z^{2} (\frac{13.6}{16}) = \frac{13.6}{4}

or

Z^{2} = 4

or

Z = 2

Q. The shortest wavelength of the Brackett series of 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

3

4

5

2

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, Z2(4213.6​)=2213.6​ or Z2(1613.6​)=413.6​ or Z2=4 or Z=2

Q. The shortest wavelength of the Brackett series of 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