Question

For the Balmer series in the spectrum of H-atom, $\mathrm{v}={\mathrm{R}}_{\mathrm{H}}\left[(1/\mathrm{n}{1}^2)-(1/\mathrm{n}{2}^2)\right]$ The correct statements among (A) to (D) are:

I) The integer ${\mathrm{n}}_1=2$

II) The ionization energy of hydrogen can be calculated from the wave number of these lines.

III) The lines of longest wavelength correspond to ${\mathrm{n}}_2=3$.

IV) As wavelength decreases, the lines of the series converge.

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

Answer 1

The correct option is C

I, III, IV

Step 1-

Explanation for correct option:

(C) I) is correct since the series studied in $\mathrm{H}$-spectrum, are de-excitation series. So, electrons get de-excited to $\mathrm{n}=2$ which means that ${\mathrm{n}}_{\mathrm{lower}}=2$.

II) we can obtain I.E. from the formula above, but since the question has been stated the formula for the Balmer series, ${\mathrm{n}}_{\mathrm{lower}}$ has been fixed as $2$. So, it is not possible to calculate I.E. from it. To calculate I.E., we’ll have to put ${\mathrm{n}}_{\mathrm{lower}}=1$, which is not possible here.

III) From the formulae,

$∆\mathrm{E}=\mathrm{hc}/\mathrm{\lambda }$
Where ${\mathrm{n}}_{\mathrm{lower}}$ fixed as $2$, $∆\mathrm{E}$ increases. So, the last line of the Balmer series, i.e. from infinity to $\mathrm{n}=2$, will have the maximum energy in the series and thus, the lowest wavelength. Similarly, For the first line in the series, i.e. from $\mathrm{n}=3$to $\mathrm{n}=2$ will have the lowest energy in the series with the highest wavelength, which makes this statement correct.

IV) As energy of a photon released on the transition from $\mathrm{n}=100$ to $\mathrm{n}=2$ will have a similar energy to that of the photon that gets released on the transition from $\mathrm{n}=101$ to $\mathrm{n}=2$ because the energy of the $100\mathrm{th}$ and the $101\mathrm{st}$ orbit will be very close in value.

In a similar way, we can see that as the higher increases, the lines start to converge together. And since increasing the higher will lead to an increase in the energy of the photon released. Which will end up releasing photons of shorter wavelengths. By Combining these two statements we can easily see that as the wavelength decreases, the spectral lines start to converge.

Explanation for Incorrect option:

(A) The above detailed explanation shows option(A) is incorrect.

(B) The above detailed explanation shows option(B) is incorrect.

(D) The above detailed explanation shows option(D) is incorrect.

Threfore, option A is correct.