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Physics Question on Atoms

According to Bohr?? theory, the wave number of last line of Balmer series is (R=1.1×107 m−1)(R = 1.1 \times 10^7 \,m^{-1})

A

5.5×105 m−15.5 \times 10^5 \,m^{-1}

B

4.4×107 m−14.4 \times 10^7 \,m^{-1}

C

2.75×106 m−12.75 \times 10^6 \,m^{-1}

D

2.75×108 m−12.75 \times 10^8 \,m^{-1}

Answer

2.75×106 m−12.75 \times 10^6 \,m^{-1}

Explanation

Solution

For last line of Balmer series, n1=2n_1 = 2 and n2=∞n_2 = \infty 1λ=R[1n12−1n22]\frac{1}{\lambda} = R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] =1.1×1074m−1=\frac{ 1.1\times10^{7}}{4} m^{-1} =2.75×106m−1 = 2.75 \times10^{6} m^{-1}