Question

Taking Rydberg's constant $R_H=1.097\times {10}^{7}m$, first and second wavelength of Balmer series in hydrogen spectrum is

• A. $2000\stackrel{{}_{\circ }}{A}$, $3000\stackrel{{}_{\circ }}{A}$
• B. $1575\stackrel{{}_{\circ }}{A}$, $2960\stackrel{{}_{\circ }}{A}$
• C. $6529\stackrel{{}_{\circ }}{A}$, $4280\stackrel{{}_{\circ }}{A}$
• D. $6552\stackrel{{}_{\circ }}{A}$, $4863\stackrel{{}_{\circ }}{A}$

Answer 1

The correct option is D $6552\stackrel{{}_{\circ }}{A}$, $4863\stackrel{{}_{\circ }}{A}$

The wavelength of photon emitted due to the transition from $n$ to $m$ state of a hydrogen atom is given by $\frac{1}{\lambda }=R_H(\frac{1}{n^2}-\frac{1}{m^2})$

Let the wavelengths of first and second line of Balmer series be ${\lambda }_1$ and ${\lambda }_2$ respectively.

Now for first wavelength of Balmer series, $n=2$ and $m=3$

$\therefore \frac{1}{{\lambda }_1}=1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{3^2})⟹{\lambda }_1\approx 6563A^o$

For second wavelength of Balmer series, $n=2$ and $m=4$

$\therefore \frac{1}{{\lambda }_2}=1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{4^2})⟹{\lambda }_2\approx 4861A^o$