MCQMediumJEE 2023Bohr's Model & Hydrogen Spectrum

JEE Physics 2023 Question with Solution

A 12.5eV12.5 \, \text{eV} electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

  • A

    22.

  • B

    11.

  • C

    33.

  • D

    44.

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: An incident electron has energy 12.5eV12.5 \, \text{eV} and bombards gaseous hydrogen.

Find: The number of spectral lines emitted.

For hydrogen, excitation from n=1n=1 to n=2n=2 requires 10.2eV10.2 \, \text{eV} and excitation from n=1n=1 to n=3n=3 requires 12.09eV12.09 \, \text{eV}. Since the incident electron energy is 12.5eV12.5 \, \text{eV}, the atom can be excited up to a maximum of n=3n=3.

The number of possible spectral lines emitted when de-excitation can occur among levels up to n=3n=3 is

3(31)2=3\frac{3(3-1)}{2}=3

Thus, the number of spectral lines emitted is 33. Therefore, the correct option is C.

Common mistakes

  • Assuming the electron can excite hydrogen to n=4n=4 is incorrect because 12.5eV12.5 \, \text{eV} is not enough for that higher excitation. First identify the maximum reachable level from the given beam energy.

  • Counting only one transition such as 323 \to 2 is wrong. Once the atom reaches a higher level, all possible downward transitions among accessible levels must be considered.

  • Using the spectral-line formula with the wrong value of nn gives an incorrect result. Use n=3n=3 as the highest excited state, then apply n(n1)2\frac{n(n-1)}{2}.

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