Single Choice

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?

A$$\lambda = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$$
Correct Answer
B$$\lambda_1 = 2\lambda_2 = 3 \lambda_3 = 4 \lambda_4$$
C$$4\lambda_1 = 2\lambda_2 = 2\lambda_3 = \lambda_4$$
D$$\lambda_1 = 2\lambda_2 = 2 \lambda_3 = \lambda_4$$

Solution

From Rydberg's formula, $$\displaystyle \frac{1}{\lambda} = Rz^2 \left ( \frac{1}{1^2} - \frac{1}{2^2} \right )$$

$$\therefore \displaystyle \lambda = \frac{4}{3Rz^2}$$

$$\lambda_1 \displaystyle = \frac{4}{3R}$$

$$\lambda_2 = \displaystyle \frac{4}{3R}$$

$$\lambda_3 = \displaystyle \frac{4}{12R}$$

$$\lambda_4\displaystyle = \frac{4}{27 R}$$
.
$$\Rightarrow \lambda_1 = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$$

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

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 ________

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

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