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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
A
$0.5 \times 10^{7}\ m^{-1}$
B
$0.25 \times 10^{7}\ m^{-1}$
C
$2.5 \times 10^{7}\ m^{-1}$
D
$0.025 \times 10^{4}\ m^{-1}$
Explanation
Series limit of Balmer: $\bar{\nu} = R/4$.
Detailed Solution
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
For the last line of the Balmer series, $n_1 = 2$, $n_2 = \infty$.
$\frac{1}{\lambda} = 10^{7}\left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{10^{7}}{4}$
$\frac{1}{\lambda} = 0.25\times10^{7}\ m^{-1}$
Note: the source paper prints the Rydberg constant as '10 $m^{-1}$'; the source solution notes it should be $10^{7}\ m^{-1}$, which is used here.
For the last line of the Balmer series, $n_1 = 2$, $n_2 = \infty$.
$\frac{1}{\lambda} = 10^{7}\left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{10^{7}}{4}$
$\frac{1}{\lambda} = 0.25\times10^{7}\ m^{-1}$
Note: the source paper prints the Rydberg constant as '10 $m^{-1}$'; the source solution notes it should be $10^{7}\ m^{-1}$, which is used here.