Atomic Structure - Result Question 102
78. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen
(1996, 1M)
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Answer:
Correct Answer: 78. $(2.725 \times 10^6 M^{-1})$
Solution:
- The Rydberg’s equation for $H$-atom is
$\dfrac{1}{\lambda}=\bar{v}(\text { wave number })=R _H\left(\dfrac{1}{n _1^{2}}-\dfrac{1}{n _2^{2}}\right)$
For Balmer series, $n _1=2$ and $n _2=3,4,5, \ldots, \infty$
For shortest $\lambda, n _2$ has to be maximum, i.e. infinity. Then
$\bar{v}=R _H\left(\dfrac{1}{4}-\dfrac{1}{\infty}\right)=\dfrac{R _H}{4}=\dfrac{1.09 \times 10^{7}}{4}=2.725 \times 10^{6} m^{-1}$
BETA
