According to Bohr’s theory, the wave number of last line of Balmer series is (R = 1.1 × 107 m-1)

Question

According to Bohr’s theory, the wave number of last line of Balmer series is (R = 1.1 × 107 m-1) (A) 5.5 × 105 m-1 (B) 4.4 × 107 m-1 (C) 2.75 × 106 m-1 (D) 2.75 × 108 m-1

Tardigrade Admin · Accepted Answer

For last line of Balmer series, n1 = 2 and n2 = ∞ (1/λ) = R[(1/n12)-(1/n22)] =( 1.1×107/4) m-1 = 2.75 ×106 m-1

Q. According to Bohr’s theory, the wave number of last line of Balmer series is $(R = 1.1 \times 1 0^{7} m^{- 1})$

$5.5 \times 1 0^{5} m^{- 1}$

$4.4 \times 1 0^{7} m^{- 1}$

$2.75 \times 1 0^{6} m^{- 1}$

$2.75 \times 1 0^{8} m^{- 1}$

For last line of Balmer series, $n_{1} = 2$ and $n_{2} = \infty$
$\frac{1}{λ} = R [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$
$= \frac{1.1 \times 1 0 ^{7}}{4} m^{- 1}$
$= 2.75 \times 1 0^{6} m^{- 1}$

Q. According to Bohr’s theory, the wave number of last line of Balmer series is (R=1.1×107m−1)

5.5×105m−1

4.4×107m−1

2.75×106m−1

2.75×108m−1