Subjective Type

The first member of the Balmer series of hydrogen atom has wavelength of $$6561\mathring { A } $$. The wavelength of the second member of the Balmer series (in $$nm$$) is ________

Solution

We know that, for hydrogen atom
$$\dfrac{1}{\lambda} = Rz^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^{2}}\right)$$

for 1st line of Balmer series, $$n_1 = 2, n_2 = 3$$

$$\dfrac{1}{\lambda_1} = Rz^2 \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)$$

$$\dfrac{1}{\lambda_1} = Rz^2\left(\dfrac{1}{4}-\dfrac{1}{9}\right)$$

$$\dfrac{1}{\lambda_1} = z^2\dfrac{5}{36}$$ ....(1)

For second line of Balmer series $$n_1 = 2, n_2 = 4$$
$$\dfrac{1}{\lambda_2} = Rz^2 \left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right)$$

$$\dfrac{1}{\lambda_2} = Rz^2\left(\dfrac{1}{4} - \dfrac{1}{16}\right)$$

$$\dfrac{1}{\lambda_2} = Rz^2 \times \dfrac{3}{16}$$ ...(2)

Dividing (1) by (2)
$$ \dfrac{\frac{1}{\lambda_1}}{\frac{1}{\lambda_2}} = \dfrac{Rz^2\frac{5}{36}}{Rz^2\frac{3}{16}} = \dfrac{5}{36} \times \dfrac{16}{3} = \dfrac{20}{27}$$

$$\dfrac{\lambda_2}{\lambda_1} = \dfrac{20}{27} \Rightarrow \lambda_2 = \dfrac{20}{27} \lambda_1$$

$$\lambda_2 = \dfrac{20}{27} \times 6561= 4860\mathring{A}$$

$$\lambda_2 = 486nm$$

SIMILAR QUESTIONS

Nuclear Physics

In a hydrogen atom, the transition takes place from $$n=3$$ to $$n=2$$. If Rydberg constant is $$1.097 \times {10}^{7} {m}^{-1}$$, the wavelength of the emitted radiation is:

Nuclear Physics

If the series limit frequency of the Lyman series is $$v_L$$, then the series limit frequency of the P fund series is?

Nuclear Physics

Hydrogen $$(_1H^1)$$, Deuterium $$(_1H^2 )$$, singly ionised Helium $$(_2He^4)^+$$ and double ionised lithium $$(_3Li^6)^{++}$$ all have one electron around the nucleus. Consider an electron transition from $$n = 2$$ to $$n=1$$. If the wave lengths of emitted radiation are $$\lambda_1, \lambda_2, \lambda_3$$ and $$\lambda_4$$ respectively then approximately which one of the following is correct?

Nuclear Physics

One of the lines in the emission spectrum of $$Li^{2+}$$ has the same wavelength as that of the $$2^{nd}$$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is $$n = 12\rightarrow n = x$$. Find the value of $$x$$.

Nuclear Physics

Given the value of Rydberg's constant is $$10^7$$ $$m^{-1}$$, the wave number of the last line of the Balmer's series in hydrogen spectrum will be:

Nuclear Physics

Hydrogen atom in ground state is excited by a monochromatic radiation of $$\lambda= 975 \overset {o}{A}$$. Number of spectral lines in the resulting spectrum emitted will be

Nuclear Physics

A hydrogen atom in its ground state is irradiated by light of wavelength 970 $$\mathring{A}$$. Taking $$hc/e = 1.237 \times 10^{-6 }$$ eV m and the ground state energy of hydrogen atom as $$-13.6$$ eV, the number of lines present in the emission spectrum is:

Nuclear Physics

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $$6561 $$$$\mathring{A}$$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

Nuclear Physics

The radius of the orbit of an electron in a Hydrogen-like atom is $$4.5 $$$$\mathrm{a}_{0}$$ where $$\mathrm{a}_{0}$$ is the Bohr radius. Its orbital angular momentum is $$\displaystyle \frac{3\mathrm{h}}{2\pi}$$ It is given that $$\mathrm{h}$$ is Planck's constant and $$\mathrm{R}$$ is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

Nuclear Physics

What would be maximum wavelength for Brackett series of hydrogen-spectrum?

Contact Details