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Atomic Structure NEET Questions — Free MCQ Practice

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The work function of a metal is 4.2 eV. If radiations of 2000 A fall on the metal then the kinetic energy of the fastest photo electron is

A.

1.6 x 10^-19J

B.

16 x 10¹⁰J

C.

3.2 x 10^-19J

D.

6.4 x 10^-10J

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Explanation

Use photoelectric effect equation

hc/ $λ$ = hc/ $λ^{0}$ + K.E ( hc/ $λ^{0}$ = work function)

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The relationship between energy E, of the radiation with a wavelength 8000 A and the energy of the radiation with a wavelength 16000 A is

A.

E₁ = 6E₂

B.

E₁ = 2E₂

C.

E₁ = 4E₂

D.

E₁ = 1/2E₂

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The tranistion of electrons in H atom that will emit maximum energy is:

A.

n₃ $\to$ n₂

B.

n₄ $\to$ n₃

C.

n₅ $\to$ n₄

D.

n₆ $\to$ n₅

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Explanation

If A is the higher level & B is the lower level then lesser is the value of B higher is the energy emitted.

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The transition in He⁺ ion in balmer series that would have the same wave number as the first Lyman line in the hydrogen spectrum is:

A.

2 $\to$ 1

B.

5 $\to$ 3

C.

4 $\to$ 2

D.

6 $\to$ 4

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Explanation

$For 1^{st} Lymen series of H = \frac{1}{λ} = {RZ}^{2} [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}] For transition of {He}^{+} in Balmer series for {He}^{+} = {RZ}^{2} \frac{1}{{(2)}^{2}} - \frac{1}{{(n_{2})}^{2}}$

$For H \frac{1}{λ_{1}} = R . {(1)}^{2} . [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}] = \frac{3 R}{4}$

$For {He}^{+} \frac{1}{λ_{2}} = R . {(2)}^{2} . \frac{1}{{(2)}^{2}} - \frac{1}{{n_{2}}^{2}} \frac{1}{λ_{2}} = 4 R \times [\frac{1}{{(2)}^{2}} - \frac{1}{{n_{2}}^{2}}]$

$Since λ_{1} = λ_{2} \frac{3 R}{4} = 4 R - [\frac{1}{4} - \frac{1}{{n_{2}}^{2}}] \frac{1}{{n_{2}}^{2}} = \frac{1}{4} - \frac{1}{16} \frac{1}{{n_{2}}^{2}} = \frac{1}{16} n_{2} = 4$

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If the speed of electron in Bohr's first orbit of hydrogen atom be x, then speed of the electron in 3rd orbit is:

A.

x/9

B.

x/3

C.

3x

D.

9x

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Explanation

$x = u_{1} = \frac{1}{1} \Rightarrow u_{1} = x F o r t h i r d o r b i t - u_{3 r d} = u_{1} \times \frac{2}{n} = x \times \frac{1}{3} = \frac{x}{3}$

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In Bohr's model of the hydrogen atom the ratio between the period of revolution of an electron in the orbit n=1 to the period of revolution of the electron in the orbit n=2 is

A.

1:2

B.

2:1

C.

1:4

D.

1:8

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Explanation

T₁/T₂ =n₁³ / n₂³ =1/8

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A cricket ball of 0.5 kg is moving with a velocity of 100 ms^-1. The wavelength associated with its motion is:

A.

1/100cm

B.

6.6 x 10^-34m

C.

1.32 x 10^-35m

D.

6.6 x 10^-28m

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Explanation

$λ = \frac{h}{m v}$ =6.6 x 10^-34/ 0.5x100 = 1.32 x 10^-35m

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How many electrons in ₁₉K have n=3; l=0?

A.

1

B.

2

C.

4

D.

3

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Explanation

₁₉K - 1S²2S²2p⁶3s²3p⁴4s¹

For 3s, n=3 & l =0

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The ratio of the difference in energy between the first and second Bohr orbit to that between the second and the third Bohr orbit is

A.

1/2

B.

1/3

C.

4/9

D.

27/5

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Explanation

$For H - atom E_{3} = - \frac{13.6}{{(3)}^{2}} = - \frac{13.6}{9} = - 1.5 eV E_{2} = - \frac{13.6}{{(2)}^{2}} = - \frac{13.6}{4} = - 3.4 eV E_{1} = - \frac{13.6}{{(1)}^{2}} = - \frac{13.6}{1} = - 13.6 eV Now, E_{2} - E_{1} = (- 3.4) - (013.6) = 13.6 - 3.4 = 10.2 eV E_{3} - E_{2} = (- 1.5) - (3.4) = 3.4 - 1.5 = 1.9 eV ∴ \frac{E_{2} - E_{1}}{E_{3} - E_{2}} = \frac{10.2}{1.9} = 5.36 \approx 5.4 = \frac{27}{5}$

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How many numbers of orbitals are possible in L-energy level?

A.

2

B.

4

C.

6

D.

1

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Explanation

L - Energy level = n=2

No. of orbitals = n² = (2)² = 4

Atomic Structure MCQs for NEET — Chemistry Questions with Answers

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