# Class 11 Chemistry MCQ – Developments Leading to the Bohr’s Model of Atom

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This set of Class 11 Chemistry Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Developments Leading to the Bohr’s Model of Atom”.

1. What’s the wavelength for the visible region in electromagnetic radiation?
a) 400 – 750 nm
b) 400 – 750 mm
c) 400 – 750 μm
d) 400 – 750 pm

Explanation: Electromagnetic spectrum is made up of various electromagnetic radiation. They are radio waves, X-rays, gamma rays, UV rays, the visible region, IR waves, and microwaves. Visible rays are the only ones which a human eye can see. They range from 450 – 750 nm.

2. What is the wavenumber of violet color?
a) 25 x 103 mm-1
b) 25 x 103 m-1
c) 25 x 103 cm-1
d) 25 x 103 nm-1

Explanation: The wavenumber is the reciprocal or the inverse of wavelength. Wavenumber = 1/Wavelength. Its unit is cm-1. The wavelength of violet color is 400nm as seen in the electromagnetic spectrum. So wavenumber = 1/400nm = 25 x 103 cm-1.

3. Calculate the frequency of the wave whose wavelength is 10nm.
a) 2 Hz
b) 3 Hz
c) 1 Hz
d) 4 Hz

Explanation: The relation between wavelength(λ) and frequency(v) of a wave is given by λ = c/v where c is the speed of light of the light. v = c/λ Frequency of the given wave = (3 x 108m/s)/(10 x 10-9m) = 3 Hz.
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4. If Energy = 4.5 KJ; calculate the wavelength.
a) 4.42 x 10-29 m
b) 4.42 x 10-39 m
c) 4.42 x 10-25 m
d) 4.42 x 10-22 m

Explanation: We know E = hv through Planck’s Quantum Theory, where E is energy, h is Planck’s constant and v is the frequency. 4.5 KJ = (6.626×10–34 Js)(3 x 108m/s)/(wavelength). wavelength = 4.42 x 10-29 m.

5. ________ frequency, is the minimum frequency required to eject an electron when photons hit the metal surface.
a) Required
b) Activated
c) Threshold
d) Limiting

Explanation: In the photoelectric effect, when photons strike on a metal surface, it emits electrons. Thus for emitting an electron, it requires a minimum amount of energy. This is threshold energy acquired through threshold frequency.

6. A metal’s work function is 3.8KJ. Photons strike metal’s surface with an energy of 5.2 KJ. what’s the kinetic energy of the emitted electrons?
a) 3.8 KJ
b) 5.2 KJ
c) 9 KJ
d) 1.4 KJ

Explanation: As per the formula of the photoelectric effect, we have E = K.E. + Wo. E is the energy of photons; K.E. is the kinetic energy with which electrons are emitted and Wo is the work function. K.E. = 5.2 KJ – 3.8 KJ = 1.4 KJ.

7. When an electron jumps from 3rd orbit to 2nd orbit, which series of spectral lines are obtained?
a) Balmer
b) Lyman
c) Paschen
d) Brackett

Explanation: As per the spectral lines of the hydrogen, when an electron jumps from nth orbit to 2nd orbit, it’s in Balmer series (provided that n = 3, 4, 5….). For Balmer series, the electron emits waves in visible region.

8. Find out the wavenumber, when an electron jumps from 2nd orbit to 1st.
a) 82357.75 cm-1
b) 105,677 cm-1
c) 82257.75 cm-1
d) 109,677 cm-1

Explanation: The Swedish spectroscopist, Johannes Rydberg gave a formula; Wavenumber = RH[(1/n1)2-(1/n2)2]. Here RH is the Rydberg constant and is equal to 109,677 cm-1. Wavenumber = 109,677(3/4) = 82257.75 cm-1.

9. The ultraviolet spectral region is obtained in Balmer series.
a) True
b) False

Explanation: When an electron jumps from nth orbit to 1st orbit, provided that n = 1, 2, 3, etc, it emits radion in the ultraviolet region. As per the spectral lines of the hydrogen, when an electron jumps from nth orbit to 2nd orbit, it’s in Balmer series (provided that n = 3, 4, 5….). For Balmer series, the electron emits waves in the visible region.

10. During the photoelectric effect, when photons strike with 5.1eV, electrons emitted from which metal have higher kinetic energy?

Metal Na Ag
Work function 2.3 eV 4.3 eV

a) Na
b) Ag
c) Equal
d) Neither

Explanation: As per the formula of the photoelectric effect, we have E = K.E. + Wo. E is the energy of photons; K.E. is the kinetic energy with which electrons are emitted and Wo is the work function. K.E. of Na and Ag are 2.8 eV and 0.8 eV.

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