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

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 (A) 2 (B) 3 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

Shortest wavelength of Brackett series corresponds to the transition of electron between

n_{1} = 4

and

n_{2} =\in f t y

and the shortest wavelength of Balmer series corresponds to the transition of electron between

n_{1} = 2

and

n_{2} =\in f t y

, So,

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

∴ Z^{2} = 4

or

Z = 2

Q. 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

$2$

$3$

$4$

$6$

Q. 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

2

3

4

6

Shortest wavelength of Brackett series corresponds to the transition of electron between n1​=4 and n2​=∈fty and the shortest wavelength of Balmer series corresponds to the transition of electron between n1​=2 and n2​=∈fty , So, (z2)(1613.6​)=(413.6​) ∴Z2=4 or Z=2

Q. 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

$2$

$3$

$4$

$6$