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Structure of Atom · Question 1

NEET (UG) 2026MCQEasy📈 Topic in 12% papers+4−1
1

A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is 4.42 × 10-19 J, how many photons are emitted by the bulb per second?

  1. 27.2 × 1019
  2. 4.06 × 1019
  3. 1.35 × 1019
  4. 2.71 × 1019

Answer

(4) 2.71 × 1019

Solution

Energy given out as light per second = 8% of 150 J = 8100 × 150 = 12 J

Number of photons per second = energy per secondenergy of one photon

⇒ n = 124.42 10-19 = 2.71 × 1019

Answer: (4)

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Keep practising — Structure of Atom

2

Match List-I with List-II :

  1. A-II, B-III, C-IV, D-I
  2. A-I, B-II, C-III, D-IV
  3. A-II, B-III, C-I, D-IV
  4. A-IV, B-II, C-III, D-I
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3

Consider the following schematic plots of orbital wavefunction ( r) against distance (r) from the nucleus. [figure] The figure representing two radial nodes in the orbital is

  1. D
  2. A
  3. B
  4. C
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4

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 n = 3 and n = 4 n = 6 transitions, respectively, is

  1. 19
  2. 14
  3. 136
  4. 116
NEET (UG) 2025
5

Energy and radius of first Bohr orbit of He+ and Li2+ are [Given RH = 2.18 10-18 J, a0 = 52.9 pm]

  1. En (Li2+ ) = -19.62 10-16 J; rn (Li2+ ) = 17.6 pm En (He+ ) = -8.72 10-16 J; rn (He+ ) = 26.4 pm
  2. En (Li2+ ) = -8.72 10-16 J; rn (Li2+ ) = 17.6 pm En (He+ ) = -19.62 10-16 J; rn (He+ ) = 17.6 pm
  3. En (Li2+ ) = -19.62 10-18 J; rn (Li2+ ) = 17.6 pm En (He+ ) = -8.72 10-18 J; rn (He+ ) = 26.4 pm
  4. En (Li2+ ) = -8.72 10-18 J; rn (Li2+ ) = 26.4 pm En (He+ ) = -19.62 10-18 J; rn (He+ ) = 17.6 pm
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6

The energy of an electron in the ground state (n = 1) for He+ ion is -x J, then that for an electron in (n = 2) state for Be3+ ion in J is

  1. -x
  2. - x9
  3. -4x
  4. - 49x
NEET (UG) 2024
7

Match List I with List II. Choose the correct answer from the options given below :

  1. A-I, B-III, C-II, D-IV
  2. A-III, B-IV, C-I, D-II
  3. A-III, B-IV, C-II, D-I
  4. A-II, B-I, C-IV, D-III
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8

The quantum numbers of four electrons are given below : I. n = 4; l = 2; ml = -2; s = - 12 II. n = 3; l = 2; ml = 1; s = + 12 III. n = 4; l = 1; ml = 0; s = + 12 IV. n = 3; l = 1; ml = -1; s = + 12 The correct decreasing order of energy of these electrons is

  1. IV > II > III > I
  2. I > III > II > IV
  3. III > I > II > IV
  4. I > II > III > IV
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9

Given below are two statements: In the light of the above statements, choose the correct answer from the options given below:

  1. Statement I is true but Statement II is false
  2. Statement I is false but Statement II is true
  3. Both Statement I and Statement II are true
  4. Both Statement I and Statement II are false
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10

Select the correct Statements from the following : Choose the correct answer from the options given below.

  1. C, D and E only
  2. A and E only
  3. B, C and E only
  4. A, B and C only
NEET (UG) 2023
11

The relation between nm (nm = the number of permissible values of magnetic quantum number (m)) for a given value of azimuthal quantum number (l), is

  1. l = 2nm + 1
  2. nm = 2l2 + 1
  3. nm = l + 2
  4. l = nm - 12
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