Hydrogen Atom Spectra Questions and Answers

Q: Different spectral lines of the Balmer series (transitions \(n \rightarrow 2\), with \(n\) being the principal quantum number) fall one at a time on a Young’s double slit apparatus. The separation between the slits is \(d\) and the screen is placed at a constant distance from the slits. What factor should \(d\) be multiplied by to maintain a constant fringe width for various lines, as \(n\) takes different allowed values?

The factor is \(\frac{4n^{2}}{n^{2} - 4}\), up to a constant.The fringe width is \(\beta = \lambda D/d\). With the screen distance fixed, keeping \(\beta\) constant requires \(d \propto \lambda\). For the Balmer series,\[ \begin{aligned} \frac{1}{\lambda} &= R\left(\frac{1}{4} - \frac{1}{n^{2}}\right) = R\,\frac{n^{2} - 4}{4n^{2}} \\ \lambda &= \frac{1}{R} \cdot \frac{4n^{2}}{n^{2} - 4} \end{aligned} \]so \(d\) must be scaled by \(4n^{2}/(n^{2} - 4)\). The option \((n^{2} - 4)/4n^{2}\) is proportional to \(1/\lambda\), the inverse of what is needed, and the forms with \(n^{2} + 4\) do not come from the Rydberg formula.

Q: The screened nuclear charge of neutral Helium atom is given as \(1.7 e\), where \(e\) is the magnitude of the electronic charge. Assuming the Bohr model of the atom for which the energy levels are \(E_{n} = -\frac{Z^{2}}{2}\frac{1}{n^{2}}\) atomic units (\(Z\) is the atomic number), the first ionization potential of Helium in atomic units is

The first ionization potential is 0.89 atomic units.In neutral helium each electron sees the screened charge \(Z = 1.7\), so with both in \(n = 1\) the atom's energy is\[ \begin{aligned} E_{\text{He}} &= 2 \times \left(-\frac{1.7^{2}}{2}\right) \\ &= -2.89 \end{aligned} \]After one electron is removed, the remaining electron in He+ sees the full nuclear charge \(Z = 2\):\[ \begin{aligned} E_{\text{He}^{+}} &= -\frac{2^{2}}{2} = -2 \\ \text{IP} &= -2 - (-2.89) = 0.89 \end{aligned} \]0.89 a.u. is about 24.2 eV, close to the measured 24.6 eV. 1.78 doubles the answer, and 1.445 (\(1.7^{2}/2\)) would be the result of removing one screened electron without letting the other relax, which is not offered.

Q: The shortest wavelength observed in Paschen back series of hydrogen spectra is (RH = 10967757·6 m−1)

The shortest wavelength is 8205 Å (the paper's 'Paschen back series' means the Paschen series).Paschen lines end on \( n = 3 \). The shortest wavelength is the series limit, an electron falling from \( n = \infty \):\[ \begin{aligned} \frac{1}{\lambda} &= R_{H}\left(\frac{1}{3^{2}} - 0\right) \\ \lambda &= \frac{9}{10967757.6}\ \text{m} \approx 8206\ \text{\AA} \end{aligned} \]The option gives this limit as 8205 Å. 7800 Å and 7349 Å are shorter than the limit, so they cannot be Paschen lines at all, and 9546 Å, longer than the limit, belongs to a transition from a finite upper level, not the shortest line.

Q: Value of radius for fifth orbital of hydrogen is (first orbital radius is 0.53 Å) :

The radius of the fifth orbit is 13.25 Å.In the Bohr model the orbit radius grows as the square of the principal quantum number, \(r_{n} = n^{2}r_{1}\):\[ \begin{aligned} r_{5} &= 5^{2} \times 0.53 \\ &= 25 \times 0.53 = 13.25\ \text{\AA} \end{aligned} \]The value 0.529 Å is the first orbit itself, and 0.26 Å and 8.4 Å do not follow from the \(n^{2}\) rule.

Q: Direct confirmation about the quantization of internal energy states of an atom was first obtained from :

Direct confirmation came from the Franck-Hertz experiment (spelt Frank-Hertz in the paper).In 1914 James Franck and Gustav Hertz accelerated electrons through mercury vapour and found that the current fell sharply whenever the accelerating voltage reached multiples of about 4.9 V. Electrons lost energy to mercury atoms only in that fixed quantum, and the vapour then emitted light at 253.7 nm. This was the first direct evidence that atoms have discrete internal energy levels.Stern-Gerlach showed space quantisation of angular momentum, Compton scattering showed the particle nature of X-rays, and Millikan's oil drop experiment measured the electron charge.

Q: The Bohr radius of the hydrogen atom and the Compton wavelength of electron are given by :

The answer is \(a \propto \frac{\hbar^{2}}{me^{2}}\) and \(\lambda_{c} \propto \frac{h}{m_{0}c}\).The Bohr radius comes from balancing the Coulomb attraction against quantised angular momentum. In Gaussian units\[ a_{0} = \frac{\hbar^{2}}{me^{2}} \]which is about 0.53 Å (in SI units \(4\pi\epsilon_{0}\hbar^{2}/me^{2}\)).The Compton wavelength is the wavelength of a photon whose energy equals the electron's rest energy:\[ \lambda_{c} = \frac{h}{m_{0}c} \]which is about \(2.43 \times 10^{-12}\) m.The other options give the wrong powers: \(\hbar^{2}/m_{0}c\) and \(\hbar/m_{0}c^{2}\) do not have the dimensions of length, and \(\hbar^{2}/mc^{2}\) has no charge in it, although the Bohr radius depends on the electric force.

Q: The wavelength of hydrogen spectrum known as Balmer series is given by :

The Balmer series is \(\frac{1}{\lambda} = \left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right)R_{H}\), n = 3, 4, 5 ...The Balmer lines are transitions that end on the n = 2 level of hydrogen. They lie in the visible region: n = 3 to 2 is the red H-alpha line at 656 nm, and the series converges at about 365 nm.The other options are the other series: lower level 1 is the Lyman series (ultraviolet), lower level 3 the Paschen series and lower level 4 the Brackett series (both infrared).

Q: The ratio of frequencies of the first line of the Lyman series and the first line of the Balmer series is :

The ratio is 27/5.By the Rydberg formula the frequency of a line is proportional to \(1/n_{1}^{2} - 1/n_{2}^{2}\). The first Lyman line is n = 2 to 1, and the first Balmer line n = 3 to 2:\[ \begin{aligned} \nu_{L} &\propto 1 - \tfrac{1}{4} = \tfrac{3}{4} \\ \nu_{B} &\propto \tfrac{1}{4} - \tfrac{1}{9} = \tfrac{5}{36} \\ \frac{\nu_{L}}{\nu_{B}} &= \frac{3}{4}\times\frac{36}{5} = \frac{27}{5} \end{aligned} \]Lyman lines are in the ultraviolet and Balmer lines in the visible, so the ratio must be greater than 1, which rules out 4/27 and 8/27. 27/8 comes from using the wrong second term for the Balmer line.

Q: If the wavelength of the first line of the Balmer series in the hydrogen spectrum is λ, then the wavelength of the first line of the Lyman series is :

The Lyman wavelength is \(\frac{5}{27}\lambda\).Use the Rydberg formula for the first line of each series:\[ \begin{aligned} \frac{1}{\lambda_{B}} &= R\left(\frac{1}{4} - \frac{1}{9}\right) \\ &= \frac{5R}{36} \\ \frac{1}{\lambda_{L}} &= R\left(1 - \frac{1}{4}\right) \\ &= \frac{3R}{4} \\ \frac{\lambda_{L}}{\lambda_{B}} &= \frac{5/36}{3/4} \\ &= \frac{5}{27} \end{aligned} \]The Lyman line is in the ultraviolet, so it must be shorter than the Balmer line; that rules out 27/5 and 32/27. 27/32 is the inverse of the ratio for a different pair of lines.

Q: A muon from cosmic rays is trapped by a proton to form a hydrogen-like atom. Given that a muon is approximately 200 times heavier than an electron, the longest wavelength of the spectral line (in analogy with the Lyman series) of such an atom will be :

The wavelength is about 6.67 Å.Energy levels of a hydrogen-like atom scale with the reduced mass. For a muon of about 200 electron masses bound to a proton (1836 electron masses), the reduced mass is about \(\frac{200 \times 1836}{2036} \approx 180\) electron masses, so every wavelength is about 180 times shorter than in hydrogen:\[ \lambda \approx \frac{1216}{180} \approx 6.7 \]in Ångström units, starting from hydrogen's Lyman-alpha line at 1216 Å.The longest Lyman-type line is the 2 to 1 transition. The nearest option is 6.67 Å. Ignoring the reduced mass (dividing by 200) gives about 6.1 Å; 13.3, 5.62 and 3.75 Å are much further off.

Different spectral lines of the Balmer series (transitions \(n \rightarrow 2\), with \(n\) being the principal quantum number) fall one at a time on a Young’s double slit apparatus. The separation between the slits is \(d\) and the screen is placed at a constant distance from the slits. What factor should \(d\) be multiplied by to maintain a constant fringe width for various lines, as \(n\) takes different allowed values?

A.\(\frac{n^{2} - 4}{4n^{2}}\)
B.\(\frac{n^{2} + 4}{4n^{2}}\)
C.\(\frac{4n^{2}}{n^{2} - 4}\)
D.\(\frac{4n^{2}}{n^{2} + 4}\)

Exam Year(s)

GATE Physics - 2023,

Answer: C

The factor is \(\frac{4n^{2}}{n^{2} - 4}\), up to a constant.

The fringe width is \(\beta = \lambda D/d\). With the screen distance fixed, keeping \(\beta\) constant requires \(d \propto \lambda\). For the Balmer series,

\[ \begin{aligned} \frac{1}{\lambda} &= R\left(\frac{1}{4} - \frac{1}{n^{2}}\right) = R\,\frac{n^{2} - 4}{4n^{2}} \\ \lambda &= \frac{1}{R} \cdot \frac{4n^{2}}{n^{2} - 4} \end{aligned} \]

so \(d\) must be scaled by \(4n^{2}/(n^{2} - 4)\). The option \((n^{2} - 4)/4n^{2}\) is proportional to \(1/\lambda\), the inverse of what is needed, and the forms with \(n^{2} + 4\) do not come from the Rydberg formula.

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The screened nuclear charge of neutral Helium atom is given as \(1.7 e\), where \(e\) is the magnitude of the electronic charge. Assuming the Bohr model of the atom for which the energy levels are \(E_{n} = -\frac{Z^{2}}{2}\frac{1}{n^{2}}\) atomic units (\(Z\) is the atomic number), the first ionization potential of Helium in atomic units is

A.0.89
B.1.78
C.0.94
D.3.16

Exam Year(s)

GATE Physics - 2025,

Answer: A

The first ionization potential is 0.89 atomic units.

In neutral helium each electron sees the screened charge \(Z = 1.7\), so with both in \(n = 1\) the atom's energy is

\[ \begin{aligned} E_{\text{He}} &= 2 \times \left(-\frac{1.7^{2}}{2}\right) \\ &= -2.89 \end{aligned} \]

After one electron is removed, the remaining electron in He⁺ sees the full nuclear charge \(Z = 2\):

\[ \begin{aligned} E_{\text{He}^{+}} &= -\frac{2^{2}}{2} = -2 \\ \text{IP} &= -2 - (-2.89) = 0.89 \end{aligned} \]

0.89 a.u. is about 24.2 eV, close to the measured 24.6 eV. 1.78 doubles the answer, and 1.445 (\(1.7^{2}/2\)) would be the result of removing one screened electron without letting the other relax, which is not offered.

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The shortest wavelength observed in Paschen back series of hydrogen spectra is (R_H = 10967757·6 m⁻¹)

A.7800 Å
B.7349 Å
C.9546 Å
D.8205 Å

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The shortest wavelength is 8205 Å (the paper's 'Paschen back series' means the Paschen series).

Paschen lines end on \( n = 3 \). The shortest wavelength is the series limit, an electron falling from \( n = \infty \):

\[ \begin{aligned} \frac{1}{\lambda} &= R_{H}\left(\frac{1}{3^{2}} - 0\right) \\ \lambda &= \frac{9}{10967757.6}\ \text{m} \approx 8206\ \text{\AA} \end{aligned} \]

The option gives this limit as 8205 Å. 7800 Å and 7349 Å are shorter than the limit, so they cannot be Paschen lines at all, and 9546 Å, longer than the limit, belongs to a transition from a finite upper level, not the shortest line.

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Value of radius for fifth orbital of hydrogen is (first orbital radius is 0.53 Å) :

A.0.529 Å
B.0.26 Å
C.8.4 Å
D.13.25 Å

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The radius of the fifth orbit is 13.25 Å.

In the Bohr model the orbit radius grows as the square of the principal quantum number, \(r_{n} = n^{2}r_{1}\):

\[ \begin{aligned} r_{5} &= 5^{2} \times 0.53 \\ &= 25 \times 0.53 = 13.25\ \text{\AA} \end{aligned} \]

The value 0.529 Å is the first orbit itself, and 0.26 Å and 8.4 Å do not follow from the \(n^{2}\) rule.

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Direct confirmation about the quantization of internal energy states of an atom was first obtained from :

A.Stern-Gerlach experiments
B.Compton scattering experiment
C.Millicon oil drop experiment
D.Frank-Hertz experiment

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

Direct confirmation came from the Franck-Hertz experiment (spelt Frank-Hertz in the paper).

In 1914 James Franck and Gustav Hertz accelerated electrons through mercury vapour and found that the current fell sharply whenever the accelerating voltage reached multiples of about 4.9 V. Electrons lost energy to mercury atoms only in that fixed quantum, and the vapour then emitted light at 253.7 nm. This was the first direct evidence that atoms have discrete internal energy levels.

Stern-Gerlach showed space quantisation of angular momentum, Compton scattering showed the particle nature of X-rays, and Millikan's oil drop experiment measured the electron charge.

Discuss in Forum

The Bohr radius of the hydrogen atom and the Compton wavelength of electron are given by :

A.\(a \propto \frac{\hbar^{2}}{me^{2}}\) and \(\lambda_{c} \propto \frac{h}{m_{0}c}\)
B.\(a \propto \frac{h^{2}}{me^{2}}\) and \(\lambda_{c} \propto \frac{\hbar^{2}}{m_{0}c}\)
C.\(a \propto \frac{h}{me}\) and \(\lambda_{c} \propto \frac{\hbar}{m_{0}c^{2}}\)
D.\(a \propto \frac{\hbar^{2}}{mc^{2}}\) and \(\lambda_{c} \propto \frac{h}{m_{0}c^{2}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The answer is \(a \propto \frac{\hbar^{2}}{me^{2}}\) and \(\lambda_{c} \propto \frac{h}{m_{0}c}\).

The Bohr radius comes from balancing the Coulomb attraction against quantised angular momentum. In Gaussian units

\[ a_{0} = \frac{\hbar^{2}}{me^{2}} \]

which is about 0.53 Å (in SI units \(4\pi\epsilon_{0}\hbar^{2}/me^{2}\)).

The Compton wavelength is the wavelength of a photon whose energy equals the electron's rest energy:

\[ \lambda_{c} = \frac{h}{m_{0}c} \]

which is about \(2.43 \times 10^{-12}\) m.

The other options give the wrong powers: \(\hbar^{2}/m_{0}c\) and \(\hbar/m_{0}c^{2}\) do not have the dimensions of length, and \(\hbar^{2}/mc^{2}\) has no charge in it, although the Bohr radius depends on the electric force.

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The wavelength of hydrogen spectrum known as Balmer series is given by :

A.\(\frac{1}{\lambda} = \left(\frac{1}{4^{2}} - \frac{1}{n^{2}}\right)R_{H}\) n = 5, 6, 7 .....
B.\(\frac{1}{\lambda} = \left(\frac{1}{3^{2}} - \frac{1}{n^{2}}\right)R_{H}\) n = 4, 5, 6 .....
C.\(\frac{1}{\lambda} = \left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right)R_{H}\) n = 3, 4, 5 .....
D.\(\frac{1}{\lambda} = \left(\frac{1}{1} - \frac{1}{n^{2}}\right)R_{H}\) n = 2, 3, 4 .....

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The Balmer series is \(\frac{1}{\lambda} = \left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right)R_{H}\), n = 3, 4, 5 ...

The Balmer lines are transitions that end on the n = 2 level of hydrogen. They lie in the visible region: n = 3 to 2 is the red H-alpha line at 656 nm, and the series converges at about 365 nm.

The other options are the other series: lower level 1 is the Lyman series (ultraviolet), lower level 3 the Paschen series and lower level 4 the Brackett series (both infrared).

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The ratio of frequencies of the first line of the Lyman series and the first line of the Balmer series is :

A.4/27
B.8/27
C.27/5
D.27/8

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The ratio is 27/5.

By the Rydberg formula the frequency of a line is proportional to \(1/n_{1}^{2} - 1/n_{2}^{2}\). The first Lyman line is n = 2 to 1, and the first Balmer line n = 3 to 2:

\[ \begin{aligned} \nu_{L} &\propto 1 - \tfrac{1}{4} = \tfrac{3}{4} \\ \nu_{B} &\propto \tfrac{1}{4} - \tfrac{1}{9} = \tfrac{5}{36} \\ \frac{\nu_{L}}{\nu_{B}} &= \frac{3}{4}\times\frac{36}{5} = \frac{27}{5} \end{aligned} \]

Lyman lines are in the ultraviolet and Balmer lines in the visible, so the ratio must be greater than 1, which rules out 4/27 and 8/27. 27/8 comes from using the wrong second term for the Balmer line.

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If the wavelength of the first line of the Balmer series in the hydrogen spectrum is λ, then the wavelength of the first line of the Lyman series is :

A.(27/32)λ
B.(27/5)λ
C.(5/27)λ
D.(32/27)λ

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The Lyman wavelength is \(\frac{5}{27}\lambda\).

Use the Rydberg formula for the first line of each series:

\[ \begin{aligned} \frac{1}{\lambda_{B}} &= R\left(\frac{1}{4} - \frac{1}{9}\right) \\ &= \frac{5R}{36} \\ \frac{1}{\lambda_{L}} &= R\left(1 - \frac{1}{4}\right) \\ &= \frac{3R}{4} \\ \frac{\lambda_{L}}{\lambda_{B}} &= \frac{5/36}{3/4} \\ &= \frac{5}{27} \end{aligned} \]

The Lyman line is in the ultraviolet, so it must be shorter than the Balmer line; that rules out 27/5 and 32/27. 27/32 is the inverse of the ratio for a different pair of lines.

Discuss in Forum

10.

A muon from cosmic rays is trapped by a proton to form a hydrogen-like atom. Given that a muon is approximately 200 times heavier than an electron, the longest wavelength of the spectral line (in analogy with the Lyman series) of such an atom will be :

A.5.62 Å
B.6.67 Å
C.3.75 Å
D.13.3 Å

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The wavelength is about 6.67 Å.

Energy levels of a hydrogen-like atom scale with the reduced mass. For a muon of about 200 electron masses bound to a proton (1836 electron masses), the reduced mass is about \(\frac{200 \times 1836}{2036} \approx 180\) electron masses, so every wavelength is about 180 times shorter than in hydrogen:

\[ \lambda \approx \frac{1216}{180} \approx 6.7 \]

in Ångström units, starting from hydrogen's Lyman-alpha line at 1216 Å.

The longest Lyman-type line is the 2 to 1 transition. The nearest option is 6.67 Å. Ignoring the reduced mass (dividing by 200) gives about 6.1 Å; 13.3, 5.62 and 3.75 Å are much further off.

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