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MCQEasyJEE 2023Bohr's Model & Hydrogen Spectrum

JEE Physics 2023 Question with Solution

A photon is emitted in transition from n=4n = 4 to n=1n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h=4×10−15 eV⋅sh = 4 \times 10^{-15} \, \text{eV} \cdot \text{s}):

  • A

    94.1 nm94.1 \, \text{nm}

  • B

    941 nm941 \, \text{nm}

  • C

    97.4 nm97.4 \, \text{nm}

  • D

    99.3 nm99.3 \, \text{nm}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Transition in hydrogen atom from n2=4n_2 = 4 to n1=1n_1 = 1. Also, h=4×10−15 eV⋅sh = 4 \times 10^{-15} \, \text{eV} \cdot \text{s} and c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}.

Find: The wavelength of the emitted photon.

For hydrogen atom, the energy difference is

E=13.6[1n12−1n22]eVE = 13.6 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \text{eV}

Substituting n1=1n_1 = 1 and n2=4n_2 = 4,

E=13.6[1−116]=13.6⋅1516E = 13.6 \left[ 1 - \frac{1}{16} \right] = 13.6 \cdot \frac{15}{16} E=12.75 eVE = 12.75 \, \text{eV}

Using $$E = \frac{hc}{\lambda}$$

Now use

E=hcλE = \frac{hc}{\lambda}

so,

λ=hcE\lambda = \frac{hc}{E}

Substituting the given values,

λ=(4×10−15)(3×108)12.75\lambda = \frac{(4 \times 10^{-15})(3 \times 10^8)}{12.75} λ=94.1 nm\lambda = 94.1 \, \text{nm}

Therefore, the wavelength of the emitted photon is 94.1 nm94.1 \, \text{nm}. Hence, the correct option is A.

The solution states "The Correct Option is D", but the working gives 94.1 nm94.1 \, \text{nm}, which matches option A. Therefore the answer is taken from the working.

Common mistakes

  • Using 1n22−1n12\frac{1}{n_2^2} - \frac{1}{n_1^2} instead of 1n12−1n22\frac{1}{n_1^2} - \frac{1}{n_2^2} for emission. This gives a negative energy difference. For emitted photon energy, take the magnitude corresponding to higher level to lower level transition.

  • Substituting the wrong quantum numbers. Here the electron goes from 44 to 11, so n2=4n_2 = 4 and n1=1n_1 = 1. Reversing them leads to incorrect evaluation of the energy gap.

  • Ignoring unit conversion in λ=hcE\lambda = \frac{hc}{E}. If hh is in eV⋅s\text{eV} \cdot \text{s} and cc is in m/s\text{m/s}, then the wavelength first comes in metres and must be converted properly to nanometres.

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