Question

For the Balmer series in the spectrum of H atom, $\bar{v} = R_{H} \left\{\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right\},$ the correct statements among (I) to (IV) are:
(I) As wavelength decreases, the lines in the series converge
(II) The integer $n_1$ is equal to 2
(III) The lines of longest wavelength corresponds to $n_2 = 3$
(IV) The ionization energy of hydrogen can be calculated from wave number of these lines

• A. (II), (III), (IV)
• B. (I), (II), (IV)
• C. (I) , ( II ) , (III)
• D. (IV), (II), (I)

Answer 1

Answer: (C)

$$
\overline{ v }= R _{ H }\left[\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right]
$$
(I) As the wavelength decreases the energy increases therefore the spectrum lines become bond and formed a converging series.
(II) For balmer series always started ground state $n _{1}=2$
(III) At longest wavelength, the higher state energy will be minimum therefore excited state will be $n _{2}=3$ nest to the ground always.
(IV) $E =13.6 \frac{ z ^{2}}{ n ^{2}}$ for $H$ aton $z =1$
So, we cannot find ionization energy directly by only ware number, we also required transition state(n) value.
According to the above observation
Only (I), (II) and (III) correct.