quantum-theory-photoelectric-effect-and-bohr-model MCQs for UPSC Prelims

29 practice questions on quantum-theory-photoelectric-effect-and-bohr-model from the Structure of Atom section of the UPSC Prelims syllabus. 29 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

7 Easy 16 Medium 6 Hard

Sample questions

Q1
medium

Assertion (A): Bohr's model cannot account for the Zeeman effect, the splitting of spectral lines in a magnetic field. Reason (R): Bohr's model assumes that the angular momentum of the orbiting electron is quantised in units of $h/2\pi$.

  1. A (a) Both A and R are true, and R is the correct explanation of A.
  2. B (b) Both A and R are true, but R is not the correct explanation of A.
  3. C (c) A is true, but R is false.
  4. D (d) A is false, but R is true.
Show answer and explanation

Correct answer: B - (b) Both A and R are true, but R is not the correct explanation of A.

Both statements are true. Splitting in a magnetic field needs states that differ in their orientation in space, an idea the Bohr model has no room for, since it allows only flat circular orbits labelled by a single number. R states one of Bohr's postulates correctly, but that postulate fixes the allowed radii and energies and says nothing about behaviour in a magnetic field, so R does not explain A.

Q2
medium

Which of the following transitions in a hydrogen atom produces a line belonging to the Paschen series?

  1. A $n = 5 \to n = 2$
  2. B $n = 3 \to n = 2$
  3. C $n = 4 \to n = 1$
  4. D $n = 5 \to n = 3$
Show answer and explanation

Correct answer: D - $n = 5 \to n = 3$

The Paschen series is made up of every transition that ends on $n = 3$, and its lines fall in the infrared. A jump ending on $n = 2$ belongs to the Balmer series in the visible region, and one ending on $n = 1$ belongs to the Lyman series in the ultraviolet. Only the jump from $n = 5$ to $n = 3$ terminates on the third level, so only it is a Paschen line.

Q3
easy

In the photoelectric effect, which statement is correct about the threshold frequency?

  1. A It is the highest frequency that can cause the emission of electrons.
  2. B It is the minimum frequency of light required to eject electrons from a metal surface.
  3. C It is the frequency above which the intensity of light no longer affects how many electrons are emitted.
  4. D It is the frequency at which all incident photons are absorbed by electrons.
Show answer and explanation

Correct answer: B - It is the minimum frequency of light required to eject electrons from a metal surface.

The threshold frequency is the minimum frequency of incident light that can eject electrons from a given metal surface; below it no electrons come out however intense the light. Above the threshold, intensity does still matter - it fixes the NUMBER of photoelectrons - so option A is false, and the threshold is a minimum rather than a maximum.

Q4
easy

According to Bohr's model, what is the radius of the second orbit (n=2) for a hydrogen atom? (Use r_n = 0.529 n^2 Å)

  1. A 4.232 Å
  2. B 0.529 Å
  3. C 1.058 Å
  4. D 2.116 Å
Show answer and explanation

Correct answer: D - 2.116 Å

Using the formula r_n = 0.529 n^2 Å, for n = 2, r_2 = 0.529 * 2^2 = 0.529 * 4 = 2.116 Å. Miscalculating n^2 or misunderstanding the formula would lead to incorrect answers.

Q5
medium

According to Planck's quantum theory, what is the energy of 5 photons of light with a frequency of 6 x 10^14 Hz? (Use Planck's constant h = 6.626 x 10^-34 J s)

  1. A 6.626 x 10^-34 J
  2. B 1.986 x 10^-19 J
  3. C 3.313 x 10^-19 J
  4. D 1.986 x 10^-18 J
Show answer and explanation

Correct answer: D - 1.986 x 10^-18 J

The energy of a single photon is given by E = hν. For multiple photons, the total energy is E_total = n * h * ν. Substituting the given values, E_total = 5 * 6.626 x 10^-34 * 6 x 10^14 = 1.986 x 10^-18 J. The distractors include incorrect multiplication or fail to multiply by the number of photons.

Q6
easy

For a hydrogen atom, what is the energy of the electron in the second orbit?

  1. A -6.8 eV
  2. B -3.4 eV
  3. C -1.51 eV
  4. D -13.6 eV
Show answer and explanation

Correct answer: B - -3.4 eV

The energy of an electron in the nth orbit is given by E_n = -13.6/n^2 eV. For n = 2, E_2 = -13.6/4 = -3.4 eV. Distractors involve incorrect squaring of n or incorrect application of the formula.

Q7
medium

Using $r_n = 0.529\,n^2\ \text{angstrom}$ for a hydrogen atom, by what factor is the radius of the $n = 4$ orbit larger than the radius of the $n = 2$ orbit?

  1. A 4
  2. B 8
  3. C 16
  4. D 2
Show answer and explanation

Correct answer: A - 4

The Bohr radius grows as $n^2$, so $r_4/r_2 = 4^2/2^2 = 16/4 = 4$. Taking the ratio as $n_2/n_1 = 2$ ignores the square, cubing the ratio instead of squaring it gives $8$, and $16$ is $4^2$ with the $n = 2$ orbit left out of the comparison. The actual radii are about $8.46$ and $2.12$ angstrom.

Q8
easy

According to Planck's quantum theory, what happens to the energy of a single quantum of radiation when the frequency of the radiation is doubled?

  1. A It is halved
  2. B It doubles
  3. C It stays the same
  4. D It becomes four times as large
Show answer and explanation

Correct answer: B - It doubles

Planck's relation $E = h\nu$ makes the energy of one quantum directly proportional to the frequency, so doubling $\nu$ doubles $E$. Halving would follow only if energy varied as $1/\nu$, which is how wavelength behaves, and a fourfold rise would need $E \propto \nu^2$. The energy cannot stay the same, because $h$ is a fixed constant and nothing else in the relation changes.

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