Question

If the short series limit of the Balmer series for hydrogen is 3646 Å. Calculate the atomic no. of the element which gives X-ray wavelength down to 1.0 Å. Identify the element –

• A. z = 21
• B. z = 31
• C. z = 11
• D. z = 5

Answer 1

Answer: (B)

z = 31

Answer 2

The short limit of the Balmer series is given by

$v = \frac { Z e ^ { 2 } } { 2 \varepsilon _ { 0 } n h }$= 1/l = R $\left( \frac { 1 } { 2 ^ { 2 } } - \frac { 1 } { \infty ^ { 2 } } \right)$ = R/4

\R = 4/l = (4/3646) × 10¹⁰ m^–1

Further the wavelengths of the K_a series are given by the relation

$r \propto n ^ { 2 }$= 1/l = R (Z – 1)²

The maximum wave number correspnods to

n =  and therfore, we must have

$r \propto n ^ { 2 }$= 1/l = R (Z – 1)² or (Z – 1)² = $\frac { 1 } { \mathrm { R } \lambda }$ = $\frac { 3646 \times 10 ^ { - 10 } } { 4 \times 1 \times 10 ^ { - 10 } }$

= 911.5 \Z – 1) =$\sqrt { 911.5 }$ $E ^ { \prime } = \frac { 36 } { 5 } E = 7.2 E$ 30.2

Or Z = 31.2 $\cong$ 31 Thus the atomic number of the element concerned is 31.The element having atomic number Z = 31 is Gallium.