Atoms and Nuclei JEE 2018: Photon Energies and Z = 3

What is the Atoms and Nuclei JEE 2018 question asking?

The answer to this Atoms and Nuclei JEE 2018 question is 3.00, not 9: the energy equation first gives the square of the atomic number. The source is JEE Advanced 2018, Paper 2, Physics, and this is a numerical-answer question, not an MCQ.

A one-electron ion emits one photon when its electron falls from level 2 to level 1, and another during a fall from level 3 to level 2. The first photon carries 74.8 eV more energy than the second. Using hydrogen’s ionisation energy of 13.6 eV, determine the atomic number.

Hydrogen-like means exactly one electron, not necessarily atomic number 1. The question bank rates this problem medium and gives an expected solving time of 120 seconds, not an official exam time limit.

How do you calculate the two emitted photon energies?

E_{n} = - \frac{13.6 Z^{2}}{n^{2}} eV

Each emitted photon has positive energy equal to the initial electron energy minus the final electron energy. The Bohr formula above gives electron energy, which is negative for a bound electron because zero is set at the ionisation limit: a free electron at rest, infinitely far from the nucleus.

During emission, the electron moves to a lower, more negative energy. For the three levels involved:

E_{1} = - 13.6 Z^{2} eV, E_{2} = - \frac{13.6 Z^{2}}{4} eV, E_{3} = - \frac{13.6 Z^{2}}{9} eV

The electron’s own energy change is negative. The photon carries away the same amount as positive energy:

Δ E_{electron} = E_{final} - E_{initial} < 0

E_{photon} = E_{initial} - E_{final} > 0

The transition symbols below denote photon energies, not changes in electron energy. For the fall from level 2 to level 1:

Δ E_{21} & = E_{2} - E_{1} & = 13.6 Z^{2} (1 - \frac{1}{4}) eV & = 13.6 Z^{2} \times \frac{3}{4} eV

For the fall from level 3 to level 2, use the same initial-minus-final order. Subtracting the negative level energies gives:

Δ E_{32} & = E_{3} - E_{2} & = 13.6 Z^{2} (\frac{1}{4} - \frac{1}{9}) eV & = 13.6 Z^{2} \times \frac{9 - 4}{36} eV & = 13.6 Z^{2} \times \frac{5}{36} eV

Calculate the two photon energies separately before comparing them. The supplied 74.8 eV is the difference between photon energies, so it cannot be assigned directly to either transition.

How does the photon-energy difference give an atomic number of 3.00?

Subtracting the second photon’s energy from the first gives 74.8 eV. This determines the square of the atomic number as 9; taking the positive square root gives 3.00. The words “74.8 eV higher” mean a difference, neither a sum nor the energy of either photon alone:

Δ E_{21} - Δ E_{32} = 74.8 eV

Substitute both expressions, with energies measured in eV. Keep the fraction subtraction in the same order:

13.6 Z^{2} (\frac{3}{4} - \frac{5}{36}) = 74.8

The common denominator is 36. Convert the first fraction before subtracting:

\frac{3}{4} = \frac{27}{36}

\frac{27}{36} - \frac{5}{36} = \frac{22}{36} = \frac{11}{18}

13.6 Z^{2} \times \frac{11}{18} = 74.8

Z^{2} = \frac{74.8 \times 18}{13.6 \times 11}

No calculator is needed. Cancel the decimal ratio first, then simplify:

\frac{74.8}{13.6} = 5.5

Z^{2} = 5.5 \times \frac{18}{11} = \frac{11}{2} \times \frac{18}{11} = 9

Atomic number is positive, so retain only the positive root. Write the numerical answer as 3.00: $Z = \sqrt{9} = 3$

Z = 3.00

Verify the answer in the original transitions, using the squared atomic number of 9. Both photons must have positive energies, and their difference must match the question:

Δ E_{21} = 13.6 \times 9 \times \frac{3}{4} = 91.8 eV

Δ E_{32} = 13.6 \times 9 \times \frac{5}{36} = 17.0 eV

91.8 - 17.0 = 74.8 eV

Why would entering 9 be wrong?

Entering 9 confuses the square of the atomic number with the atomic number itself. This question has no answer options, so 9 is a possible incorrect numerical response, not an official distractor.

Bohr energies scale with the square of the atomic number. Reaching the following line and then entering 9 leaves the calculation unfinished: $Z^{2} = 9$ $Z = \sqrt{9} = 3$

The second line is the required square-root step. Substituting an atomic number of 9 would instead square it to 81:

13.6 \times 81 \times \frac{11}{18} = 673.2 eV

That is nine times the stated difference. Keep the requested variable visible, and substitute your final value into the original photon-energy difference before entering the answer.

Which three practice questions check the same Bohr-model method?

These original practice questions are not additional verified JEE PYQs. They check substituting the atomic number, recovering it from its square, and cancelling a shared factor in a ratio. Attempt each before reading its worked check.

For singly ionised helium, find the photon energy emitted from level 2 to level 1.

Helium has atomic number 2. The ion’s single positive charge does not change its nuclear atomic number.

{H e}^{+} : Z = 2

Δ E_{21} = 13.6 \times 4 \times \frac{3}{4} = 40.8 eV

A one-electron ion emits a 17.0 eV photon from level 3 to level 2. Find its atomic number.

Here, the given energy belongs to one photon, not a difference between photons. Use that transition alone, then take the square root.

13.6 Z^{2} \times \frac{5}{36} = 17

Z^{2} = \frac{17 \times 36}{13.6 \times 5} = 9, Z = 3

For any hydrogen-like ion, what is the ratio of the photon energies from level 2 to level 1 and level 3 to level 2?

Both transitions belong to the same ion. Their shared squared-atomic-number factor cancels, so the ratio does not require identifying the element.

\frac{Δ E_{21}}{Δ E_{32}} = \frac{13.6 Z^{2} (3 / 4)}{13.6 Z^{2} (5 / 36)} = \frac{3 / 4}{5 / 36} = \frac{27}{5} = 5.4

Now reproduce the original solution without looking. Include the two positive photon energies, the fraction subtraction and the final square root.

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If that step was the hard part, work through How to Study Kinetic Theory JEE: Formulas and Practice.

Frequently asked questions

What is the answer to the Atoms and Nuclei JEE Advanced 2018 question?

The atomic number is 3, entered as 3.00. The photon-energy difference gives 13.6 × Z² × (11/18) = 74.8, so Z² = 9. Taking the positive square root gives Z = 3.

Why is 9 the wrong answer in this JEE 2018 question?

The energy equation gives Z² = 9, not Z = 9. Entering 9 skips the final square-root step. Substituting Z = 9 gives a photon-energy difference of 673.2 eV instead of the stated 74.8 eV.

How do you calculate photon energy for a hydrogen-like ion?

Use the Bohr electron-energy formula E_n = −13.6Z²/n² eV. For emission, photon energy equals the initial electron energy minus the final electron energy, so it is positive. The 2-to-1 and 3-to-2 transitions give 13.6Z² × 3/4 eV and 13.6Z² × 5/36 eV, respectively.

Does hydrogen-like mean the atomic number is 1?

No: hydrogen-like means the atom or ion has exactly one electron. Its atomic number can be greater than 1. For example, singly ionised helium has one electron but an atomic number of 2.

Atoms and Nuclei JEE 2018: Bohr Model Solution