Question

According to Bohr’s theory, the wave number of last line of Balmer series is $(R = 1.1 \times 10^7 \,m^{-1})$

• A. $5.5 \times 10^5 \,m^{-1}$
• B. $4.4 \times 10^7 \,m^{-1}$
• C. $2.75 \times 10^6 \,m^{-1}$
• D. $2.75 \times 10^8 \,m^{-1}$

Answer 1

Answer: (C)

For last line of Balmer series, $n_1 = 2$ and $n_2 = \infty$
$\frac{1}{\lambda} = R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] $
$=\frac{ 1.1\times10^{7}}{4} m^{-1} $
$ = 2.75 \times10^{6} m^{-1}$