First slide

Hydrogen -like ion

Question

Ratio of shortest wavelength of Balmer series in a hydrogen atom to that of Lymann series in a $H e^{+}$ ion is

Moderate

A

1 : 4
B

8 : 1
C

1 : 8
D

16 : 1

Solution

for any series shortest wavelength can be given by $\frac{1}{λ} = R Z^{2} (\frac{1}{n^{2}} - \frac{1}{\infty}) w h e r e R i s t h e R y d b e r g' s c o n s \tan t a n d Z i s t h e a t o m i c n o .$
For Balmer series in hydrogen atom $\frac{1}{λ_{1}} = R (\frac{1}{2^{2}} - \frac{1}{\propto}) = \frac{R}{4}$
For Lymann series of $H e^{+}$ is $\frac{1}{λ_{2}} = 4 R (\frac{1}{1^{2}} - \frac{1}{\infty})$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{4 / R}{1 / 4 R} = 16$

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