Question 595: Atoms & Nuclei

Question

The wavelength of the first line of Lyman series for 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

Accepted Answer

The wavelength of the first line of lyman
series for hydrogen atom is
$${1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$
The wavelength of the second line of Balmer series
for hydrogen like ion is
$${1 \over {\lambda '}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$
According to question
$$\lambda $$ = $$\lambda $$'
$$R\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$ = $$R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$
$$ \Rightarrow $$ $${3 \over 4}$$ = $${{3{Z^2}} \over {16}}$$
$$ \Rightarrow $$ Z = 2

Answer

3

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4

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1

Answer

2

Atoms & Nuclei

The wavelength of the first line of Lyman series for 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

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