First slide

Hydrogen -like ion

Question

A hydrogen – like ion having wavelength difference between first line of Balmer and Lymann series is $593 \overset{o}{A} .$ Then atomic number of the ion is _____

Moderate

A

1
B

2
C

3
D

4

Solution

$\begin{array}{l} \frac{1}{λ_{1}} = z^{2} R (1 - \frac{1}{2^{2}}) = \frac{3 R Z^{2}}{4} \\ \frac{1}{λ_{1}} = z^{2} R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R Z^{2}}{36} \\ N o w, λ_{2} - λ_{1} = \frac{36}{5 R Z^{2}} - \frac{4}{3 R Z^{2}} = \frac{88}{15 R Z^{2}} \\ ∴ \frac{88}{Z^{2} (15 \times 1.097 \times 10^{7})} = 593 \times 10^{- 10} \\ \Rightarrow Z = 3. \end{array}$

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