Atoms Ques 6
6. Given the value of Rydberg constant is $10^{7} m^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be : [2016]
(a) $0.025 \times 10^{4} m^{-1}$
(b) $0.5 \times 10^{7} m^{-1}$
(c) $0.25 \times 10^{7} m^{-1}$
(d) $2.5 \times 10^{7} m^{-1}$
Show Answer
Answer:
Correct Answer: 6.(c)
Solution:
- (c) According to Bohr’s theory, the wave number of the last line of the Balmer series in hydrogen spectrum,
For hydrogen atom $Z=1$
$\frac{1}{\lambda}=RZ^{2}(\frac{1}{n_2^{2}}-\frac{1}{n_1^{2}})$
$=10^{7} \times 1^{2}(\frac{1}{2^{2}}-\frac{1}{\infty^{2}})$
$\Rightarrow$ wave number $\frac{1}{\lambda}=0.25 \times 10^{7} m^{-1}$
Last line of the series is called series limit. For this line wavelength is minimum $(\lambda _{\min })$.
For minimum wavelength, $n_2=\infty, n_1=n$
So, wavelength $\lambda _{\min }=\frac{n^{2}}{R}$
BETA
