MCQEasyJEE 2026Bohr Model & Hydrogen Spectrum

JEE Chemistry 2026 Question with Solution

The energy required by electrons, present in the first Bohr orbit of hydrogen atom, to be excited to second Bohr orbit is \hspace{1.5cm} J mol1^{-1}.

Given: RH=2.18×1011R_H = 2.18 \times 10^{-11} ergs.

  • A

    9.835×10129.835 \times 10^{12}

  • B

    9.835×1059.835 \times 10^{5}

  • C

    1.635×10111.635 \times 10^{-11}

  • D

    1.635×10181.635 \times 10^{-18}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Electrons are excited from the first Bohr orbit to the second Bohr orbit in hydrogen atom. Also, RH=2.18×1011R_H = 2.18 \times 10^{-11} ergs.

Find: The energy required for excitation and the correct option.

The energy of an electron in the nthn^{th} Bohr orbit is

En=RHn2E_n = -\frac{R_H}{n^2}

For the first and second orbits,

E1=RH,E2=RH4E_1 = -R_H, \quad E_2 = -\frac{R_H}{4}

Hence, the energy required for excitation is

ΔE=E2E1=(RH4)(RH)=3RH4\Delta E = E_2 - E_1 = \left(-\frac{R_H}{4}\right) - (-R_H) = \frac{3R_H}{4}

Substituting the given value,

ΔE=34×2.18×1011=1.635×1011\Delta E = \frac{3}{4} \times 2.18 \times 10^{-11} = 1.635 \times 10^{-11}

Therefore, the required energy is 1.635×10111.635 \times 10^{-11}, so the correct option is C.

The solution states the final answer as 1.635×1011J mol11.635 \times 10^{-11} \, \text{J mol}^{-1}, although the given constant is written in ergs; this unit inconsistency is present in the source.

Common mistakes

  • Using E1E2E_1 - E_2 instead of E2E1E_2 - E_1 without interpreting excitation properly. For excitation, the electron moves to a higher orbit, so the required energy is the increase in energy. Always calculate the final-state energy minus the initial-state energy.

  • Ignoring the negative sign in Bohr energy levels. Both orbit energies are negative, and the difference must be taken carefully. Write E1=RHE_1 = -R_H and E2=RH4E_2 = -\frac{R_H}{4} explicitly before subtracting.

  • Substituting the orbit number incorrectly as n=2n = 2 directly into the formula for the whole transition. The formula En=RHn2E_n = -\frac{R_H}{n^2} gives energy of one orbit, not the excitation energy. First find both orbit energies, then take the difference.

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