NEET UG PHYSICS MCQ: ATOMS

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Maximum wavelength in Balmer series of hydrogen spectrum is (R : Rydberg constant)

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Correct option: D

\frac{{36}}{R}

Option D (

\frac{{36}}{R}

) is correct because the maximum wavelength in the Balmer series corresponds to the transition from n=3 to n=2. Using the Rydberg formula,

\frac{1}{\lambda} = R(\frac{1}{2^2} - \frac{1}{3^2}) = R(\frac{1}{4} - \frac{1}{9}) = R(\frac{5}{36})

. Therefore,

\lambda = \frac{36}{5R}

.

More Questions from ATOMS

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Which of the following statements are true regarding Bohr’s model of hydrogen atom

(I) Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus

(II) Radii of allowed orbits of electron are proportional to the principal quantum number

(III) Frequency with which electrons orbit around the nucleus is discrete orbits is inversely proportional to the cube of principal quantum number

(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits

Select correct answer using the codes given below
A particle moving with a velocity of

\frac{1}{{100}}\mathrm{th}

of that of light will cross a nucleus on about A hydrogen atom is in ground state absorbs a photon of ultraviolet radiation of wavelength

50 \mathrm{~nm}

. Assuming that the entire photon energy is taken up by electron with what KE will the electron be emitted. In a spectrum of hydrogen atom, the ratio of shortest wavelength in Lyman series to longest wavelength in Paschen series is When the wave of hydrogen atom comes from infinity into the first orbit then the value of wave number is To explain his theory, Bohr used