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

z2((13 . 6/16))=(13 . 6/4) ⇒ z2=4⇒ 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

$2$

$3$

$4$

$6$

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

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

z2(1613.6​)=413.6​ ⇒z2=4⇒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

$2$

$3$

$4$

$6$

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