# The wavelength of the first line of Lyman series for the hydrogen atom is equal to that of the second line of Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is. Option 1) 3 Option 2) 4 Option 3) 1 Option 4) 2

As we learnt in

Lymen series -

$\frac{1}{\lambda }= R\left ( \frac{1}{1^{2}}- \frac{1}{n^{2}} \right )$

$\left ( uv \: region \right )$

- wherein

$n=2,3,4,-----$

When electron jump from higher orbital to 1st energy level

i.e. ground state

AND

Balmer series -

$\frac{1}{\lambda }= R\left ( \frac{1}{2^{2}}- \frac{1}{n^{2}} \right )$

$\left ( \: visible \: light \right )$

- wherein

$n=3,4,5-----$

When electron jump from higher orbital to n=2 energy level

Lymen series of hydrogen

$\frac{1}{\lambda } = R \left ( \frac{1}{1^{2}}-\frac{1}{2^{2}} \right ) \: \: = \frac{3R}{4}$

Second line of Balmer series

$\frac{1}{\lambda } = R Z^{2} \left ( \frac{1}{4}-\frac{1}{16} \right ) \: \: = \frac{3R}{16}Z^{2}$

According to question,

$\frac{3R}{4}= \: \frac{3R Z^{2}}{16} \: \: \: \: \Rightarrow Z^{2}=4 \ \: or \: \: Z=2$

Option 1)

3

Incorrect

Option 2)

4

Incorrect

Option 3)

1

Incorrect

Option 4)

2

Correct

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