Question 21: Atoms & Nuclei

Question

Identify the hydrogen-like element whose spectral lines are four times shorter in wavelength compared to those of atomic hydrogen.

Accepted Answer

For hydrogen-like atom,
$${1 \over {\lambda '}} = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$ ..... (i)
Similarly, for hydrogen atom,
$${1 \over \lambda } = R\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$ ....... (i)
For the element to have hydrogen-like spectrum, the ratio of the wavelengths for transition of electron between any two levels n
1
and n
2
should be constant.
Dividing eq. (ii) by eq. (i), we get
$${{\lambda '} \over \lambda } = {{R\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)} \over {R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)}}$$ or $${{\lambda '} \over \lambda } = {1 \over {{Z^2}}}$$
It is given that wavelength of the hydrogen atom  is four times the wavelength of the hydrogen-like atom, i.e.
$$\lambda  = 4\lambda '$$
Hence, $$Z = 2$$
Now, the element with atomic number 2 is helium.

Answer

Lithium

Answer

Helium

Answer

Berilliyum

Answer

Potassium

Atoms & Nuclei

Identify the hydrogen-like element whose spectral lines are four times shorter in wavelength compared to those of atomic hydrogen.

Choose the correct answer: