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Q.

The shortest wavelength of Lyman series of hydrogen is equal to the shortest wavelength of Balmer series of a hydrogen-like atom of atomic number Z. The value of Z is

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a

2

b

6

c

3

d

4

answer is B.

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Detailed Solution

$\begin{array}{l} {(λ_{min})}_{Lymen} \to \begin{array}{c} n_{1} = 1 \\ n_{2} = \infty \end{array}\} z_{1} = 1 \\ {(λ_{\min})}_{Balmer} \to \begin{array}{c} n_{1} = 2 \\ n_{2} = \infty \end{array}\} z_{2} = ? \\ \frac{1}{λ} = R (z)^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) ∴ z^{2} α \frac{1}{(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})} \\ ∴ {(\frac{z_{1}}{z_{2}})}^{2} = \frac{\frac{1}{2^{2}} - \frac{1}{\infty}}{\frac{1}{1^{2}} - \frac{1}{\infty}}; ∴ {(\frac{1}{z^{2}})}^{2} = \frac{1}{4}; ∴ z_{2} = 2 \end{array}$

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