Atoms and Nuclei JEE 2023: Atomic Number Problem Solved

What is the Atoms and Nuclei JEE 2023 question and its answer?

The answer to this atoms and nuclei jee 2023 problem is 3: the photon carries 5.95 eV, not the electron’s 1.95 eV. The source is JEE Advanced 2023, Paper 1, Physics, not JEE Main.

A hydrogen-like, one-electron atom emits radiation when its electron falls from the fourth level to the third. This radiation ejects electrons from a metal. The maximum photoelectron kinetic energy is 1.95 eV, and the metal’s threshold wavelength is 310 nm. Find the atom’s atomic number using:

h c = 1240 e V n m, R h c = 13.6 e V .

This is a numerical-answer question, rated medium on the question bank’s scale. Its bank-listed target time is 120 seconds, not an official time limit.

How do you calculate the photon energy for the fourth-to-third-level transition?

Subtract the final atomic energy from the initial atomic energy to get the positive emitted photon energy. Keep the atomic number symbolic at this stage, following the official solution’s order. “Hydrogen-like” means the species has one electron, not necessarily one proton; its nuclear charge is: $+ Z e .$

The Bohr energy formula and the two relevant levels are:

E_{n} = - \frac{13.6 Z^{2}}{n^{2}} e V .

E_{4} = - \frac{13.6 Z^{2}}{16} e V, E_{3} = - \frac{13.6 Z^{2}}{9} e V .

The third level is more negative, so the atom loses energy during the transition. The atom’s energy change is negative, but the photon carries away a positive amount:

Δ E_{a t o m} = E_{3} - E_{4} < 0, h ν = E_{4} - E_{3} > 0 .

Substitute, retaining both minus signs before simplifying:

h ν = - \frac{13.6 Z^{2}}{16} - (- \frac{13.6 Z^{2}}{9}) = 13.6 Z^{2} (\frac{1}{9} - \frac{1}{16}) e V .

h ν = 13.6 Z^{2} (\frac{16 - 9}{144}) = 13.6 Z^{2} (\frac{7}{144}) e V .

Do not assign this expression to the photoelectron’s kinetic energy. It represents the full transition photon energy, before any energy is spent escaping the metal.

How does the 310 nm threshold wavelength give a photon energy of 5.95 eV?

The threshold wavelength gives the metal’s work function, which is 4 eV. It is a property of the metal, not the wavelength of the incident transition radiation. At threshold, a photon just frees an electron, leaving zero maximum kinetic energy.

ϕ = \frac{h c}{λ_{0}} = \frac{1240 e V n m}{310 n m} = 4 e V .

The nanometre units cancel, leaving electronvolts. Apply Einstein’s photoelectric equation: $K_{max} = h ν - ϕ .$

1.95 e V = h ν - 4 e V .

h ν = (1.95 + 4) e V = 5.95 e V .

Keep these three quantities separate:

Photon energy, 5.95 eV: total energy delivered by one incident photon.
Escape energy, 4 eV: the metal’s work function.
Maximum electron kinetic energy, 1.95 eV: what remains after escape.

For practice with this energy balance, use How to Study Dual Nature JEE: Six Steps and Practice.

How do you solve for the atomic number and verify 3?

Equating the two photon-energy expressions gives an atomic number of 3. Use the full 5.95 eV photon energy, with energies expressed in eV:

5.95 = 13.6 Z^{2} (\frac{7}{144}) .

Z^{2} = \frac{5.95 \times 144}{13.6 \times 7} .

Z^{2} = \frac{856.8}{95.2} = 9 .

Atomic number counts protons and must be positive, so take the positive root: $Z = \sqrt{9} = 3 .$

Check both stages by substitution:

E_{γ} = 13.6 \times 9 \times \frac{7}{144} = 5.95 e V .

K_{max} = 5.95 - 4 = 1.95 e V .

An atomic number of 3 with only one electron identifies the ion: ${L i}^{2 +} .$

It is not neutral lithium. Atomic number counts protons; ionic charge reflects the imbalance between protons and electrons.

Final numerical entry: 3.

Why does using 1.95 eV as the photon energy give the wrong answer?

Using 1.95 eV as the photon energy omits the energy needed to escape the metal. This is a numerical-answer question, and the supplied record contains no answer options. The calculation below is a wrong numerical route, not a supplied distractor.

The incorrect equation, with energies in eV, is:

1.95 = 13.6 Z^{2} (\frac{7}{144}) .

Carrying that error through gives:

Z^{2} = \frac{1.95 \times 144}{13.6 \times 7} = \frac{280.8}{95.2} \approx 2.95 .

$Z \approx 1.72 .$ Rounding this result to 2 does not repair the model. The incident photon must supply both the escape energy and the outgoing electron’s kinetic energy:

Transition photon energy = Work function + Maximum kinetic energy .

A non-integer atomic number should trigger a check of the setup, not automatic rounding. Repair the energy accounting before repeating the algebra.

How do you solve two related atomic-spectrum questions?

Use the same transition-energy method for both exercises, then either convert energy into wavelength or compare it with the metal’s work function. These are original practice variations, not additional verified JEE PYQs. Attempt each before reading its worked answer, using the supplied constants:

h c = 1240 e V n m, R h c = 13.6 e V .

Question 1, original practice variation: For the same one-electron ion with atomic number 3, find the energy and wavelength of the photon emitted in the transition from the third level to the second.

The positive photon energy is the initial atomic energy minus the final atomic energy:

E_{γ} = 13.6 \times 9 (\frac{1}{4} - \frac{1}{9}) = 13.6 \times 9 \times \frac{5}{36} = 17 e V .

Convert that photon energy into wavelength:

λ = \frac{h c}{E_{γ}} = \frac{1240}{17} n m \approx 72.94 n m .

Question 2, original practice variation: A hydrogen-like ion with atomic number 2 emits a photon in the fourth-to-third-level transition. Can it eject electrons from the original metal, whose threshold wavelength is 310 nm?

Calculate the photon energy and compare it with the escape cost:

E_{γ} = 13.6 \times 4 \times \frac{7}{144} \approx 2.64 e V, ϕ = \frac{1240}{310} = 4 e V .

No photoemission occurs, because the photon cannot supply the work function. A negative result from subtracting the work function means no emission, not electrons emerging with negative kinetic energy. Check this inequality before calculating any photoelectron speed.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

Is this atoms and nuclei question from JEE Main or Advanced 2023?

This numerical-answer question is from JEE Advanced 2023, Paper 1, Physics, not JEE Main. It combines a hydrogen-like atom’s fourth-to-third-level transition with photoelectric emission from a metal.

How do you find the atomic number in this JEE Advanced 2023 question?

The metal’s work function is 1240/310 = 4 eV, so the incident photon energy is 4 + 1.95 = 5.95 eV. Equating this to the Bohr transition energy gives 5.95 = 13.6Z²(1/9 − 1/16), hence Z² = 9. Atomic number is positive, so the final numerical answer is 3.

Why is the photon energy 5.95 eV instead of 1.95 eV?

The 1.95 eV value is the maximum kinetic energy of the emitted electron, not the incident photon energy. The photon must also supply the metal’s 4 eV work function, making its total energy 5.95 eV. Using 1.95 eV in the transition formula gives an incorrect atomic number of about 1.72; rounding it does not fix the error.

Why is the atom Li²⁺ and not neutral lithium?

Atomic number 3 means the nucleus contains three protons. The question specifies a hydrogen-like species, which has only one electron, so its net charge is +2 and it is Li²⁺. Neutral lithium has three electrons and is not the one-electron species described.